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sdas [7]
3 years ago
15

What is bigger3/4 or 7/8

Mathematics
1 answer:
JulsSmile [24]3 years ago
3 0

\frac{3}{4 }  <  \frac{7}{8}

\frac{3}{4}  = 0.75

\frac{7}{8}  = 0.875

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66.64 rounded to the nearest cent
Zolol [24]
It would just be 66.6
4 0
3 years ago
Navy PilotsThe US Navy requires that fighter pilots have heights between 62 inches and78 inches.(a) Find the percentage of women
Zigmanuir [339]

The first part of the question is missing and it says;

Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.

Answer:

A) Percentage of women meeting the height requirement = 72.24%

B) Percentage of men meeting the height requirement = 0.875%

C) Corresponding women's height =67.42 inches while corresponding men's height = 72.19 inches

Step-by-step explanation:

From the question,

For men;

Mean μ = 68.6 in

Standard deviation σ = 2.8 in

For women;

Mean μ = 63.7 in

Standard deviation σ = 2.9 in

Now let's calculate the standardized scores;

The formula is z = (x - μ)/σ

A) For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.7224

So, percentage of women meeting the height requirement is 72.24%.

B) For men;

Z = (62 - 68.6)/2.8 = -2.36

Z = (78 - 68.6)/2.8 = 3.36

Thus, the probability of only men will take the form of;

P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961

And P(Z > -2.36) = 0.99086

Thus;

P(Z<3.36) - P(Z > -2.36) 0.99961 - 0.99086 = 0.00875

So, percentage of women meeting the height requirement is 72.24%.

B)For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875

So, percentage of women meeting the height requirement is 0.875%

C) Since the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women.

For women;

Let's find the z-value with a right-tail of 10%. From the second table i attached ;

invNorm(0.90) = 1.2816

Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches

For men;

We have seen that,

invNorm(0.90) = 1.2816

Thus ;

Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches

7 0
4 years ago
The life spans of elephants in a Columbus zoo are normally distributed. The average elephant lives 60 years: the standard deviat
Marta_Voda [28]

did u ever get the answer l need it


5 0
3 years ago
How do I solve this literal equation? <br> 15 points
kati45 [8]

Answer:

\pi = \frac{A}{2r^2 + 2rh}

Step-by-step explanation:

First, we need to isolate \pi by taking it common from both terms on the right:

A=2\pi r^2 + 2\pi rh\\A=\pi (2r^2 + 2rh)

Now, since we want \pi in terms of the other variables, we can divide the left hand side (A) by whatever is multiplied with \pi on the right hand side. Then we will have an expression for \pi. Shown below:

A=\pi (2r^2 + 2rh)\\\pi = \frac{A}{2r^2 + 2rh}

This is the xpression for \pi

7 0
3 years ago
Find the measure of the marked acute angle to the nearest degree
Setler79 [48]
Answer: 62°
Solution:
tan x =34/18
x=tan^-1 (34/18)
x=62°
5 0
4 years ago
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