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3 years ago

4) Mr. Conner is starting a marching band at his school. Mr. Conner has 210 students who are interested in

Mathematics

1 answer:

zhenek [66]3 years ago

6 0

Answer:168

Step-by-step explanation:80% × 210= 0.80× 210 = 168

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At the beginning of the first day, Ann has 25 dollars. By the end of the first day, Ann has 2/3 of the money that she had at the

ira [324]

Answer:

Ann have at the end of the second day $\$9\frac{11}{21}.$

Step-by-step explanation:

Given:

At the beginning of the first day, Ann has 25 dollars.

By the end of the first day, Ann has 2/3 of the money that she had at the beginning of the first day.

By the end of the second day, Ann has 4/7 of the money that she had at the end of the first day.

Now, to find the money Ann have at the end of the second day.

Ann has at the beginning of the first day = $25.

At the end of the first day Ann had:

$\frac{2}{3} \ of\ 25$

$=\frac{2}{3} \times 25$

$=\$\frac{50}{3} .$

So, at the end of the first day Ann had $=\$\frac{50}{3} .$

Now, to get the money Ann have at the end of the second day:

$\frac{4}{7} \ of\ \frac{50}{3}$

$=\frac{4}{7} \times \frac{50}{3}$

$=\frac{200}{21}$

$=\$9\frac{11}{21} .$

Therefore, Ann have at the end of the second day $\$9\frac{11}{21}.$

8 0

3 years ago

Someone please help me

SVEN [57.7K]

V=πr^2 h/3
V=π(3)^2 (4)/3
V=37.7 yd^3 or 36 yd^3

3 0

2 years ago

What is the exact area of a semicircle with diameter of 28 inches?

erastova [34]

The answer is 307.72

$(\pi \times {14}^{2} )/2$

3 0

2 years ago

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How many lines of symmetry doesn’t a regular polygon with 32 sides have?

notsponge [240]

A regular polygon of "n" sides has "n" Lines of Symmetry

so a regular 32 sided polygon has 32 lines of symmetry

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2 years ago

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

8 0

3 years ago