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Lubov Fominskaja [6]

3 years ago

7

The amount of money Blain earns as a service technician after x hours is modeled by the equation y = $20x + 100. What is the mea

ning of the $20 in this model?

1 answer:

Margarita [4]3 years ago

3 0

The $20 in the model means that Blain works for $20 dollars per hour

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I need help with this problem y=5x-3

defon

Answer:

Slope= 5

y intercept= (0,-3)

Step-by-step explanation:

3 0

3 years ago

hram777 [196]

Answer:

D: x < 20

Step-by-step explanation:

-4x + 16 > -64

subtract 16 from both sides because you must isolate x.

-4x + 16 - 16 > -64 - 16

-4x > -80

Then divide -4 on both sides... I believe that when you divide a number by a negative number you must flip the sign...

-4x/ -4 < -80/ -4

x < 20

(negative divided by a negative will result in a POSITIVE.)

Hope this helped... I truly apologize if this is wrong..

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3 years ago

Grandma is making a

Svetlanka [38]

Answer:

the answer is 2304

Step-by-step explanation:

8×8=64

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4 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

F/4 + 16 < 49<br> Solve the inequality

Mnenie [13.5K]

Answer:

$f$ < $132$

Step-by-step explanation:

$Greetings!$

$I'm~Isabelle~Williams~and~I~will~be~answering~your~question!$

$Subtract~16~from~both~sides:$

$\frac{f}{4} +16-16$ < $49-16$

$Simplify:$

$\frac{f}{4}$ < $33$

$Now,~multiply~both~sides~by~4:$

$\frac{4f}{4}$ < $33\times4$

$f$ < $132$

$Hope~my~answer~helps!~Have~a~great~day~ahead!$

$-Isabelle~Williams$

5 0

4 years ago