Simplify it and please show work

Answer:

1014 in

Step-by-step explanation:

(13 x 13) x 6 = 1014 in^3

Answer:

The answer should be A (676)

Step-by-step explanation:

Answer:

8 5 7 1 3 2 4 6

Step-by-step explanation:

my Cashsap kinqcollin

Ewan HH

Step-by-step explanation:

Ewanjajsbehudbeiisiejebejehehehheehhehehe

Answer:

???

Step-by-step explanation:

The total animals = 139, and we can substitute this value into equation (3) to obtain:x = 2(139) - 279 => x = -1This doesn't make sense since we can't have negative chickens.

Let the number of chickens be x and the number of cows be y in the backyard. We can create two equations based on the given information:The total number of animals is x + y = total animals ...

(1)The total number of eyes and legs can also be expressed as 2x + 2y = 366 (since each animal has two eyes) and 2x + 4y = 558 (since each chicken has 2 legs and each cow has 4 legs).

Simplifying the equations, we get:x + y = total animals ... (1)x + 2y = 279 ... (2)From equation (1),

we can substitute y = total animals - x in equation (2) to obtain:x + 2(total animals - x) = 279 => 2total animals - x = 279 => x = 2total animals - 279 ... (3)We also know that x + y = total animals, so y = total animals - x.

Substituting equation (3) into this expression, we get:y = total animals - (2total animals - 279) => y = 279 - total animals ... (4)

To find the value of x, we need to know the value of total animals. We can obtain this value by adding equations (1) and (4):x + y + (279 - total animals) = total animals => x + y = 279/2 => x + y = 139.5

Since x and y are both whole numbers, they must add up to an even number. The only possible value is 70 + 69 = 139.  Therefore, there must be a mistake in the problem statement or calculations.

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Answer:

119

Step-by-step explanation:

A=w 1=7x17=119

Answer:

The sample does not provide sufficient evidence that women today are taller than women in 1990.

The difference in mean height is not significant to arrive at such a conclusion.

Step-by-step explanation:

Mean height of women 20 year of age or older in 1990 = 63.7 inches

Mean height of women 20 years of age or older in 2015 = 63.9 inches

The standard deviation of the mean height in 2015 = 0.5 inch

The difference between the two years' mean height is 0.2 (63.9 - 63.7) inch.

This difference is less than the standard deviation of the mean height in 2015.

Therefore, the difference does not provide sufficient evidence that women today are taller than women in 1990.

Answer: AA

Step-by-step explanation:

You can do vertical angles or alternate interior angles. Hope this helps :)

Answer:

A) The parameter of interest is the average filling weight of all cereal packages. ;

Point estimate = 1.22

Margin of Error = 0.0196

Confidence interval = (1.2004, 1.2396)

C) No, since the confidence interval contains the target filling weight of 1.20.

Step-by-step explanation:

A) The parameter of interest is the average filling weight of all cereal packages.

The point estimate = sample mean = 1.22

The margin of error with 95% confidence :

Zcritical * σ/sqrt(n)

Zcritical at 95% confidence = 1.96

Margin of Error = 1.96 * 0.06/sqrt(36)

Margin of Error : 1.96 * 0.01

Margin of Error = 0.0196

Confidence interval :

xbar ± margin of error

1.22 ± 0.0196

Lower boundary = 1.22 - 0.0196 = 1.2004

Upper boundary = 1.22 + 0.0196 = 1.2396

(1.2004, 1.2396)

C) No, since the confidence interval contains the target filling weight of 1.20.

Since, the internal contains 1.20, then we can conclude that it is filling properly.

Answer:

F

(

−

2

x

+

12

)

=

−

8

x

3

+

144

x

2

−

864

x

+

1731

Step-by-step explanation:

Answer:

4/7

Step-by-step explanation:

The probability that a student does not play an instrument given that they play a sport is 4/23.

It is the chance of an event to occur from a total number of outcomes.

The formula for probability is given as:

Probability = Number of required events / Total number of outcomes.

We have,

Total number of students = 23

From the Venn diagram,

Number of students who play sport but not an instrument = 4

The probability that a student does not play an instrument given that they play a sport.

= 4/23

Thus,

The probability is 4/23.

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Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that [tex]\mu = 100, \sigma = \sqrt{36} = 6[/tex]

Sample of 25:

This means that [tex]n = 25, s = \frac{6}{\sqrt{25}} = 1.2[/tex]

(a) P(X<99)

This is the pvalue of Z when X = 99. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{99 - 100}{1.2}[/tex]

[tex]Z = -0.83[/tex]

[tex]Z = -0.83[/tex] has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{100 - 100}{1.2}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

X = 98

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{98 - 100}{1.2}[/tex]

[tex]Z = -1.67[/tex]

[tex]Z = -1.67[/tex] has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

Answer:

Cylinder

Step-by-step explanation:

A cone is a 3D triangle with a circular base, and a sphere is a 3D circle

Answer:

CYLIDER

Step-by-step explanation:

Answer:

54ft

Step-by-step explanation:

9ft x 6ft

Answer:

28.27

Step-by-step explanation:

3.14 x 9

is how to do this problem

Answer:

I want her mollah

Step-by-step explanation:

Answers:

d = v(initial)t + 1/2(at^2)

v(final) = √(v(initial)^2 + 2ad)

a = 2(d-v(initial)t) / t^2

t = -v(initial)±√(v(initial)^2 + 2ad) / a

Answer:

1. vt-d=1/2at^2

2. d-vt=1/2at^2

3. a=2(d-vt)/t^2

Step-by-step explanation:

Answer:

third option

Step-by-step explanation:

subtract 125/4 from each side to get:

x² + 11x -1

use quadratic formula to get:

-11 ± √121 - 4(-1)/2

= -11 ± ([tex]\sqrt{125}[/tex]) / 2

* [tex]\sqrt{125}[/tex] = [tex]\sqrt{25}[/tex] · [tex]\sqrt{5}[/tex] = 5[tex]\sqrt{5}[/tex]

= -11 ± (5[tex]\sqrt{5}[/tex]) / 2

Answer:

....

Step-by-step explanation:

mean: 16

median: 16

there is no mode

range: 8

Answer:

LDM

MDL

Step-by-step explanation:

The names of a figure are gotten from the points on the figure. The names of the figure are LDM  and MDL.

There are three points on the given figure, and these points are: point L, point D and point M, where Point D is between L and M.

This means that, when naming the figure, alphabet D must be at the middle while alphabets M and L can be at either sides of D.

So, the possible names of the figure are: LDM  and MDL.

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Answer:

secured loan

Step-by-step explanation:

Answer: Here

Step-by-step explanation:

You Go

Answer:

The wingspan of a pelican is 1 8/15 feet longer than the wingspan of an eagle.

(1 whole 8/15 feet)

Step-by-step explanation:

We'll need to find a common denominator for the fractions. 15 is the lowest number divisible by both 5 and 3, so we'll use 15. For this problem, it's also easier to convert to improper fractions, in my opinion.

Pelican:

8 1/5 = 41/5

41/5 x 3/3 = 123/15

Eagle:

6 2/3 = 20/3

(20/3) x (5/5) = 100/15

Find the difference:

123/15 - 100/15 = 23/15 = 1 8/15

Please let me know if you have questions.

[tex]6^{2}+8^{2}=x^{2} \\36+64=x^{2} \\100=x^{2} \\x=10[/tex]

Answer:

3x+5-7

Step-by-step explanation:

Answer:

If im not mistaken it should be answer 3. Because you would have to divide amongst groups.

Answer:

[tex]3\frac{7}{8}[/tex]

Step-by-step explanation:

Equation: 4 1/7 + 2 1/8 + x= 10 1/7

Convert into improper fractions:

29/7 + 17/8= 71/7

Add fractions

351/56+x= 71/7

Subtract on both sides

X= 31/8 or 3 7/8

Answer:

The average function [tex]f_{avg}[/tex] value = 4

Step-by-step explanation:

Given that:

[tex]f(x,y,z) = 3x^2z+3y^2z[/tex]

The enclosed region [tex]z = 4 -x^2 -y^2[/tex] and the plane z = 0.

Over an area E, the average value of a function f(x,y,z) is:

[tex]f_{avg} = \dfrac{1}{V_{E}} \iiint_E f(x,y,z) dV ------ (1)[/tex]

where;

[tex]V_E[/tex] = volume of region E

To find the volume use cylindrical coordinates;

[tex]x=rcos \theta; \ \ \ y = rsin \theta; \ \ \ z = z[/tex]

Then;

E = 0 ≤ r ≤2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4 - r²

[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} \int^{4-r^2}_{0} \ rdzdrd \theta[/tex]

[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} r(4-r^2) \ drd \theta[/tex]

[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} (4r-r^3) \ drd \theta[/tex]

[tex]V_E = \int^{2 \pi}_{0} \Big(2r^2- \dfrac{r^4}{4} \Big)^2_0 \ d \theta[/tex]

[tex]V_E = 4(2 \pi)\\\\ V_E = 8 \pi[/tex]

However, from equation (1):

[tex]f_{avg} = \dfrac{1}{V_E} \iiint_E f(x,y,z) \ dV \\ \\ = \dfrac{1}{8 \pi} \iiint_E (3x^2z + 3y^2 z) dV \\ \\ = \dfrac{1}{8 \pi } \int^{2 \pi}_{0} \int^{2 }_{0} \int^{4 - r^2}_{0} \ \ 3zr^2 \ rdzdrd\theta \\ \\ = \dfrac{3}{8 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} \Big(\dfrac{z^2}{2}\Big)^{4-r^2}_{0} \ r^3 drd\theta \\ \\ = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (4-r^2)^2 \ r^3 \ dr d\theta[/tex]

[tex]= \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (16-8r^2 + r^4)r^3 \ dr d\theta \\ \\ = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (16r^3 -8r^5 + r^7) \ dr d \theta \\ \\ = = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \Big (4r^4 - \dfrac{8}{6}r^6 + \dfrac{r^8}{8} \Big)^2_0 \ d \theta \\ \\ = \dfrac{3}{16 \pi } (2 \pi) \Big (64 - \dfrac{8}{6} \times 64 \times \dfrac{256}{8} \Big) \\ \\ = \dfrac{3}{16 \pi} (2 \pi) (\dfrac{32}{3}) \\ \\ = 4[/tex]

Answer:

3x+18+4x+15=180

7x+33=180

7x=147

x=21