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3 years ago

If 5x-8w=9, then solve for x, in terms of w and z

Mathematics

2 answers:

babymother [125]3 years ago

7 0

Answer:

Step-by-step explanation:

5x-8w=9 add 8w : 5x-8w+8w = 9+8w

5x = 9+8w

now divid by 5 : x = (9+8w)/5

Alex Ar [27]3 years ago

5 0

X=7 w=3 it’s a guess and check to find out which of the characters fit and just find a difference if 9 between the two

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Solve the inequality<br> -2(y+5)+22>12-2y

zhannawk [14.2K]

Answer:

There are no solutions

Step-by-step explanation:

Always remember to simplify both sides of the inequality.

$-2y+12>-2y+12$

Add 2y to both sides.

$-2y+12+2y>-2y+12+2y\\12>12$

Subtract 12 from both sides.

$12-12>12-12\\0>0$

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3 years ago

Darien could swim 30 laps an

fredd [130]

Answer:

Step-by-step explanation:

If the percent change is 30 to 36 you can already see that it is an increase

here is the work

36 - 30

x 100% = 20%

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2 years ago

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Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or alwa

andrew-mc [135]

Answer:

We are given that 81% of respondents said that their jobs were sometimes or always stressful.

So, p = 0.81

(a) Approximate the probability that 150 or fewer workers find their jobs stressful.

x = 150

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{150}{175} =\widehat{p}$

$0.8571=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8571-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=1.5882$

Refer the z table

So, P(z<150)=0.9429

So, the probability that 150 or fewer workers find their jobs stressful is 0.9429

b) Approximate the probability that more than 140 workers find their jobs stressful.

x = 140

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{140}{175} =\widehat{p}$

$0.8=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=-0.3372$

P(z<140) = 0.3707

P(z>140)=1-P(z<140)=1-0.3707=0.6293

So, the probability that more than 140 workers find their jobs stressful is 0.6293

c) Approximate the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive.

x = 146

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{146}{175} =\widehat{p}$

$0.8342=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8342-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=0.816$

From part a) P(z<150)=0.9429

So,P(z<150)-P(z<146)=0.9429 - 0.8342=0.1087

Hence the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive is 0.1087

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3 years ago

The weight of

Feliz [49]

10 000 samples wiegh 1 pound.

1 sample would weigh (1 / 10 000) pound

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3 years ago

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Mademuasel [1]

Answer:

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Step-by-step explanation:

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3 years ago

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