Please help I need the answer ASAP!!

The rectangular field has a perimeter of 98 meters. The width of the perimeter is three times as much as five times the length.

White raven [17]

Answer:

a) x = 3.0625m

y = 45.9375 m

b) A = 140.68359375 m²

Step-by-step explanation:

Let x be the length

Let y be the width

2x + 2y = 98

x + y = 49

y = 3(5x)

y = 15x

x + 15x = 49

16x = 49

x = 3.0625 m

y = 15(3.0625) = 45.9375

A = xy = 3.0625(45.9375) = 140.68359375

4 0

3 years ago

Second pictures of my last question

Elanso [62]

Answer:

you really did not give us the full pictuer

Step-by-step explanation:

8 0

2 years ago

How do I solve this?

Snowcat [4.5K]

M>B = 65 so arc AC = 2(65) = 130
arc ABC = 360 - 130 = 230

m<D = 1/2(230 - 130)
m<D = 1/2(100) = 50

answer
m<D = 50 first choice

6 0

3 years ago

Could use help with this one! Will give thanks and brainliest to correct answer

cluponka [151]

I believe the answer is A

5 0

3 years ago

Read 2 more answers

Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would a

Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

6 0

3 years ago