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Vinvika [58]
3 years ago
7

HELP PLEASE :) * giving brainliest*

Mathematics
2 answers:
Molodets [167]3 years ago
8 0

Answer:

(2,16)

Step-by-step explanation:

Step 1: Substitute -x+18 for y in y=8x:

-x+18=8x <em>Add X to both sides</em>

9x=18 <em>Divide 9 by both sides</em>

x=2

y=8(2)<em>Plug in the new value for x(which is 2) into one of the equations</em>

y=16

(2,16)

Hope this helped, good luck!

Inessa05 [86]3 years ago
5 0

Answer:

(2,16)

Step-by-step explanation:

y=-x+18

y=8x

8x=-x+18

8x+x=9x

-x+x=0

9x=18

9x/9=x

18/9=2

x=2

y=8(2)

y=16

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Answer:

The rearrangement can be 45, 987 , 310

Step-by-step explanation:

Here, we want to rearrange the number such that 9 is worth 10 times as what it is worth presently

The value of 9 presently is 90,000

So 10 times as worth will be 10 * 90,000 = 900,000

So we can have the new arrangement as;

45, 987, 310

3 0
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A principal of $950 is invested in an account at 7 percent simple interest per year. What is the amount of the principal after 5
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nirvana33 [79]

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Step-by-step explanation:

$54 divided by 9= $6

5 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
3 years ago
What is the value of x
nordsb [41]
Median is half of add the sum ofthe base and first side
17+5x-14=3+5x
divide it by 2
3+5x ÷2=median
(3+5x)/2=x+12
3+5x=2x+24
3x=21
x=7

Answer is D=7

8 0
3 years ago
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