Translate the algebraic expression below into a written expression.

Mathematics

1 answer:

Vinvika [58]4 years ago

3 0

The absolute value of the quantity of five times x added to nine.
This will be answer .

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An employee worked 36 hours last week. This week, she worked 48 hours. Find the percent increase. Round to the nearest percent.

Ratling [72]

Answer:

To the nearest percentage, percent increase is 33%

Step-by-step explanation:

Mathematically;

Percentage increase = (new - old)/old * 100%

new = 48 hours

old = 36 hours

% increase = (48-36)/36 * 100%

12/36 * 100%

1/3 * 100%

= 33.33%

6 0

3 years ago

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A bag contains 4 white, 3 black, and 6 green balls. Balls are picked at random. Explain why the events picking a white ball and

Fofino [41]

Answer: D

Step-by-step explanation:

D is the correct answer because first of all, there are 13 balls total. So the denominator would have to be 13. Also, white is the second largest number of balls in the set, so it is pretty likely that you will pick it. Although the most common picked would be green since it has the most balls.

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3 years ago

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The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95