Answer:
2/5
Step-by-step explanation:
Multiply all the fraction
Solve the equation. 5 = g/8 what does g equal
Answer:
Step-by-step explanation:
multiply both sides by 8
do over and write g=8x5
solve 8x5=40
g=40
Pls help i'm struggling with this
Answer:
Click the second one, I truly hope that worked! :)
Find surface area plzzzzz
Answer:
184 m²
Step-by-step explanation:
Surface Area = 2(10*2) + 2(10*6) + 2(2*6)
Surface Area = 2(20 + 60 + 12)
Surface Area = 2(92)
Surface Area = 184 m²
If my answer is incorrect, pls correct me!
If you like my answer and explanation, mark me as brainliest!
-Chetan K
solve kx-3=5 for x
a
b
c
or
d
Answer:
x = 8/k
Step-by-step explanation:
kx - 3 = 5
Add 3 to each side
kx -3+3= 5+3
kx = 8
Divide by k
kx/k = 8/k
x = 8/k
Answer:
x = 8 / k
Step-by-step explanation:
Let's solve for x.
kx − 3 = 5
Step 1: Add 3 to both sides.
kx −3 +3 = 5 + 3
kx = 8
Step 2: Divide both sides by k.
kx / k = 8k
x = 8 / k
area of the triangle ABC
Answer:
180
Step-by-step explanation:
hope this answer be right :)
help me solve this pleaseeee
Answer:
f(n) = 6n + 12
Step-by-step explanation:
There is a common difference in consecutive number of seats, that is
42 - 36 = 36 - 30 = 30 - 24 = 24 - 18 = 6
This indicates the sequence is arithmetic with nth term
[tex]a_{n}[/tex] = a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 18 and d = 6 , then
f(n) = 18 + 6(n - 1) = 18 + 6n - 6 = 6n + 12
if g(x) = 3x + 7 what is the value of g(-2)?
1- -1
2- 1
3- 8
4- 12
Answer:
12
Step-by-step explanation:
3x+7 = 12
A rectangular park is 260 m by 480 m. Danny usually trains by running 5 circuits around the edge of
the park. After heavy rain, two adjacent sides are too muddy to run along, so he runs a triangular path
along the other two sides and the diagonal. Danny does 5 circuits of this path for training. Show that
Danny runs about 970 metres less than his usual training session
Answer:
971 m to nearest metre
Step-by-step explanation:
training distance = 5*2*(260 + 480) m
Diagonal distance = √(2602 + 4802) m
Heavy rain training distance = 5*(260 + 480 + √(2602 + 4802)) m
Difference between these:
5 × 2 × ( 260 + 480 ) - 5 ×( 260 + 480 + √ 260 ^ 2 + 480 ^ 2 )
= 970 . 5311872087638744
difference = 971 m to nearest metre
Hope this answer helps you :)
Have a great day
Mark brainliest
My mama got me doing freaking school work pls help me out
Answer:
x=50
Step-by-step explanation:
corresponding angles
The owner of a small restaurant bought 75 kilograms of rice. Each week, the restaurant uses 4.5 kilograms of rice.
Function r gives the remaining amount of rice, in kilograms, as a function of the number of weeks since the restaurant
owner bought the rice.
Complete the table. Type the answers in the boxes below.
WEEKS
KILOGRAMS OF RICE LEFT
6
12
Answer:
6 weeks = 48 kg
12 weeks = 21 kg
Step-by-step explanation:
6 weeks:
4.5 x 6 = 27 kg
75 - 27 = 48 kg
12 weeks:
4.5 x 12 = 54 kg
75 - 54 = 21 kg
Hope this helps!
If the farm has 10,000 spinach plants, Describe a method that would produce a random sample and how many in your sample study?
Answer:
Take an even number of random samples from the 10000 spinach plants
lets assume you take ; 100 random samples each sample will contain = 10,000 / 100 = 100 spinach plants
Step-by-step explanation:
Given that the total number of spinach plants in the farm = 10,000
To obtain a random sample, take a census of the number of farms wherein the spinach plants are planted, determine the mean size and standard deviation of the farm that way you can apply the central limit theorem.
After determining these Take an even number of random sample from the 10000 spinach plants
lets assume you take ; 100 random samples each sample will contain = 10,000 / 100 = 100
Find the shaded sector and the length of the arc. Round to the nearest hundredth.
Answer:
Step-by-step explanation:
I'll show you how to do the first one; the other are exactly the same, so pay attention.
The formula for arc length is
[tex]AL=\frac{\theta}{360}*2\pi r[/tex] where θ is the central angle's measure. It just so happens that the measure of the central angle is the same as the measure of the arc it intercepts. Our arc shows a measure of 40°; this measure is NOT the same as the length. Measures are in degrees while length is in inches, or cm, or meters, etc. Going off that info, our central angle measures 40°. Filling in the formula and using 3.1415 for π:
[tex]AL=\frac{40}{360}*2(3.1415)(3)[/tex]. I'm going to reduce that fraction a bit (and I'll use the same reduction in the Area of a sector coming up next):
[tex]AL=\frac{1}{9}*2(3.1415)(3)[/tex] which makes
AL = 2.09 units. Now for Area of the Sector. The formula is almost identical, but instead uses the idea that the area of a circle is πr²:
[tex]A_s=\frac{\theta}{360}*\pi r^2[/tex] where θ is, again, the measure of the central angle (which is the same as the measure of the arc it intercepts). Filling in:
[tex]A_s=\frac{1}{9}*(3.1415)(3)^2[/tex] which simplifies a bit to
[tex]A_s=\frac{1}{9}*(3.1415)(9)[/tex]. As you can see, the 9's cancel each other out, leaving you with
[tex]A_s=3.14[/tex] units²
Answer:
Step-by-step explanation:
[tex]Arc \ Length = 2 \pi r \frac{ \theta}{360}[/tex]
[tex]Sector \ Area = \pi r^2 \frac{ \theta }{360}[/tex]
[tex]1) Arc \ Length = 2 \pi \times 3 \times \frac{40}{360} = 2\pi \times 3 \times \frac{1}{9} = \frac{2}{3} \pi = 2.09[/tex]
[tex]Sector \ Area = \pi \times 9 \times \frac{40}{360} = 9\pi \times \frac{1}{9} = \pi = 3.14[/tex]
[tex]2) Arc \ Length = 2 \pi \times 5 \times \frac{88}{360}= 7.68[/tex]
[tex]Sector \ Area = \pi \times 25 \times \frac{88}{360} = 19.2[/tex]
[tex]3) Arc \ Length = 2 \pi \times 6 \times \frac{260}{360} = 27.22[/tex]
[tex]Sector \ Area = \pi \times 36 \times \frac{260}{360} = 81.68[/tex]
PLEASE PLEASE PLEASE SOMEONE HELP ME THIS IS DUE TODAY
EXPLAINING YOUR ANSWER = BRAINLIEST
What is m∠E?
Answer:
try 144
Step-by-step explanation:
m<E is basically just ot asking you to find the angle of E, i think :').
And since all angles equal to 180, 180-36=144
(hope ots right)
Find WX. Assume that segments which appear to be tangent are tangent.
Answer:
Step-by-step explanation:
Tangent from the point outside the circle are equal
WX = XY
7x - 29 = 2x + 16 {Add 29 to both sides}
7x = 2x + 16 +29
7x = 2x + 45 {Subtract 2x from both sides}
7x - 2x = 45
5x = 45 {Divide both sides by 5}
x = 45/5
x = 9
WX = 7x - 29
= 7*9 - 29
= 63 - 29
WX = 34
Any of you guys know this answer?
Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims that there are rotations about the center that will carry both a regular hexagon and a regular nonagon (9-sided polygon) onto itself. Determine whether each angle of rotation below can be used to support Nadia's claim. Select Yes or No for each angle of rotation
Answer:
[tex]\theta_1 \ n\ \theta_2 = 120, 240[/tex]
Step-by-step explanation:
The question is incomplete, as the angles of rotation are not stated.
However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself
We have:
[tex]Nonagon = 9\ sides[/tex]
[tex]Hexagon = 6\ sides[/tex]
Divide 360 degrees by the number of sides in each angle, then find the multiples.
Nonagon
[tex]\theta = \frac{360}{9} =40[/tex]
List the multiples of 40
[tex]\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320[/tex]
Hexagon
[tex]\theta = \frac{360}{6} =60[/tex]
List the multiples of 60
[tex]\theta_2 = 60, 120, 180, 240, 300[/tex]
List out the common angles
[tex]\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320[/tex]
[tex]\theta_2 = 60, 120, 180, 240, 300[/tex]
[tex]\theta_1 \ n\ \theta_2 = 120, 240[/tex]
This means that, only a rotation of [tex]120, 240[/tex] will lift both shapes onto themselves, when applied to both shapes.
The other angles will only work on one of the shapes, but not both at the same time.
write an equation of the line in slope intercept form
Answer:
[tex]y=-2x+2[/tex]
Step-by-step explanation:
[tex]y=mx+b[/tex] is the slope intercept
[tex]m[/tex] is the slope
[tex]b[/tex] is the y-intercept
The slope in the graph is 4/2 or 2 simplified. From point (-2,6), you have to go down 4 and right 2 to get to the point (0,2)
The y-intercept is 2 because the line crosses the y axis at exactly 2.
Hope this helps
(140-x)/(7)+70 = 120. Please answer correctly.
Adam is refinishing his basement. 30 ft by 50 ft are the dimensions of Adam's rectangular basement. How much tile would he need to cover the floor? If the height of the basement is 6 ft, how much wall area is there to paint?
Answer:
Area of four walls is 3960 square feet.
Step-by-step explanation:
length, L = 30 ft
Width, W = 50 ft
height, H = 6 ft
The area of walls is
= 2 (L W + W H + H L)
= 2 (30 x 50 + 50 x 6 + 6 x 30)
= 2 (1500 + 300 + 180)
= 3960 square feet
Answer/Step-by-step explanation:
How much tile would he need to cover the floor?
Answer: 1500 Ft^2
If the height of the basement is 6 ft, how much wall area is there to paint?
Answer: 960 Ft^2
Help a bab out with something, please?
Answer:
18
Step-by-step explanation:
What is the amplitude in the graph of y = 5sin(2x - 1) + 3?
Answer:
The amplitude is 5
Step-by-step explanation:
The function [tex]y=asin(bx-c)+d[/tex] has an amplitude of [tex]|a|[/tex]. Therefore the amplitude of the function [tex]y=5sin(2x-1)+3[/tex] is [tex]|a|=|5|=5[/tex].
Put these numbers in order from least to greatest.
0.78
8.7
0.078
Answer:
Least:0.078
In between: 0.78
Greatest: 8.7
Step-by-step explanation:
Two lines are intersecting. What is the value of X? And to your answer in the box
Answer:
x = 51
Step-by-step explanation:
The two angles form a straight line
Angles that form a straight line add up to equal 180
Hence, 2x + 4 + x + 23 = 180
( note that we just created an equation that we can use to solve for x )
we now solve for x using the equation we created
2x + 4 + x + 23 = 180
step 1 combine like terms
2x + x = 3x
4 + 23 = 27
we now have 180 = 27 + 3x
step 2 subtract 27 from each side
180 - 27 = 153
27 - 27 cancels out
we now have 153 = 3x
step 3 divide each side by 3
153 / 3 = 51
3x / 3 = x
we're left with x = 51
Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone (Prasad et al., 2000). Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. The mean overall duration of symptoms for the zinc lozenge group was 4.5 days, and the standard deviation of overall duration of symptoms was 1.6 days. For the placebo group, the mean overall duration of symptoms was 8.1 days, and the standard deviation was 1.8 days.
a. Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the zinc lozenges.
b. Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the placebo lozenges.
c. On the basis of the intervals computed in parts (a) and (b) and/or the picture you constructed in part (c), is it reasonable to conclude generally that taking zinc loz-enges reduces the overall duration of cold symptoms more than if taking a placebo? Explain why you think this is or is not an appropriate conclusion.
d. In their paper, the researchers say that they checked whether it was reasonable to assume that the data were sampled from a normal curve population and decided that it was. How is this relevant to the calculations done in parts (a) and (b)?
Answer:
a. (3.83952, 5.16049)
b. (7.32, 8.88)
c. The difference in the range of the confidence intervals suggest that taking the zinc loz-enges reduces the overall duration of cold symptoms
d. The assumptions required for the validity validity of a t-test include that the data is sampled from a source that have data that are normally distributed
Step-by-step explanation:
The given parameters are;
The number of participants that took zinc lozenges, n₁ = 25
The number of participants that took a placebo, n₂ = 23
The mean duration of the symptoms for the zinc lozenge group, [tex]\overline x_1[/tex] = 4.5 days
The standard deviation of overall duration, s₁ = 1.6 days
The mean duration of the symptoms for the placebo group, [tex]\overline x_2[/tex] = 8.1 days
The standard deviation, s₂ = 1.8 days
a. The 95% confidence interval if everyone in the population were to take zinc lozenges is given as follows;
[tex]CI=\bar{x}_1\pm t_{\alpha/2} \times \dfrac{s_1}{\sqrt{n_1}}[/tex]
n₁ = 25, the degrees of freedom (df) = n₁ - 1 = 24
The t-value for 95% confidence interval, with df = 24, with t = 2.064
Therefore, we have;
[tex]CI_{zl}=4.5 \pm 2.064 \times \dfrac{1.6}{\sqrt{25}}[/tex]
[tex]CI_{zl}[/tex] = 4.5 ± 0.66048
[tex]CI_{zl}[/tex] = (3.83952, 5.16049)
b. The 95% confidence interval if everyone in the population were to take the placebo is given as follows;
[tex]CI_p=\bar{x_2}\pm t_{\alpha/2} \times \dfrac{s_2}{\sqrt{n_2}}[/tex]
n₂ = 23, the degrees of freedom (df) = n₂ - 1 = 22
The t-value for 95% confidence interval, with df = 22, with t = 2.074
Therefore, we have;
[tex]CI_p =8.1 \pm 2.074 \times \dfrac{1.8}{\sqrt{23}}[/tex]
[tex]CI_p[/tex] = 8.1 ± 0.7784
[tex]CI_p[/tex] = (7.32, 8.88)
c. Based on the difference in the range of the 95% confidence interval for the mean duration of cold symptoms of the group that take zinc lozenges, (3.83952, 5.16049), and the mean of the group that are on a placebo, (7.32, 8.88), there is sufficient statistical evidence to suggest that there is a difference between the mean and taking zinc lozenges reduces the overall duration of cold symptoms more than if taking placebo
d. The assumptions required for validity when carrying out a t-test include;
i) The measurement scale is ordinal or continuous
ii) The sample is a simple random sample
iii) The data gives a normal distribution curve
iv) The sample size is reasonably large
v) The variance are homogeneous
Among these assumptions, the data must be randomly sampled from the population of interest and that the data variables follow a normal distribution
The common assumptions made when doing a t-test include those regarding the scale of measurement, random sampling, normality of data distribution, adequacy of sample size and equality of variance in standard deviation
Can you please help me with this
B
5
6
100° 5
A
Ta
60°
4 120°
W
D
w = [? 1°
Answer:
I'm sorry but this is very wordy and hard for people to understand.
please reconsider fixing it and submitting it again for assistance.
Question: A sample contains 68 grams of radioactive Bruinium. Sixteen days later, the sample had decayed to 47 grams.
A) what is the decay rate, k?
B) what is half-life of bruinium to the nearest 0.1 day?
C) how much bruinium would be left in another 10 days? (After the first 16)
D) how much time is required for the sample to decay to 25 grams (from t=0)?
In the diagram, the measure of angle 3 is 105°.
A transversal intersects 2 lines to form 8 angles. Clockwise from the top, the angles are 1, 2, 3, 4; 5, 6, 7, 8.
Which angle must also measure 105°?
Angle1
Angle4
Angle6
Angle8
Answer:
angle 8
Step-by-step explanation:
not sure but if correct 1 heart pls and comment
Answer: A is Correct
Step-by-step explanation:
angle 1 is opposite of angle 3 which means its also 105°
The table of values represents an exponential function f(x). What is the average rate of change over the interval −2≤x≤2? Enter your answer, as a decimal rounded to the nearest hundredth, in the box.
−3 64
−2 16
−1 4
0 1
1 1/4
2 1/16
3 1/64
(Worth 60pts)
Answer:
-1456
-864
4
-11
-5,25
-1,93
Step-by-step explanation:
Answer:
-3.98Step-by-step explanation:
Rate of change is:
[f(2) - f(-2)]/[2 - (-2)] = (1/16 - 16)/4 = -255/16*1/4 = -255/64 = -3.981. Two identical cubes each of total surface area of 6 cm2 are joined end to end. Which of the following is the total surface area of the cuboid so formed?
Answer:
10 cm³
Step-by-step explanation:
The total surface area of a cube = 6 cm²
The formula for the surface area of the cube is :
A = 6a², a is the side of the cube
6 = 6a²
a = 1 cm
Now, the length of the formed cuboid = 2 cm
Breadth = 1 cm
Height = 1 cm
Required volume of the formed cubiod is :
V = 2(lb+bh+hl)
Put all the values,
V = 2(2(1)+1(1)+2(1))
V = 10 cm³
So, the required volume of the formed cuboid is equal to 10 cm³