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3 years ago

There are 90 chairs set up in the gym for an assembly. After the assembly, all the chairs will

1 answer:

6 0

12 racks because 90 chairs divided into 8 racks, 11 racks is fully stacked with 8 chairs within it. while an addion rack will be added because 2 chairs is left. so the answer is 12 racks

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What are zeros of the function? What are there multiples? f(x) = 4x^3 -20x2 + 24x

bixtya [17]

When X=0, the function would be:

<span>f(x) = 4x^3 -20x2 + 24x
0= </span><span>4x^3 -20x2 + 24x ----->divide all by x
</span>x(4x^2 -20x + 24) =0 ------> split -20x into -12x and -8x
x(4x^2 -12x -8x + 24)
x{4x(x-3) - 8(x -3}
x(4x-8) (x-3)
x1= 0
x2= 8/4= 2
x3= 3

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3 years ago

Use the diagram shown to find the height of the house. Round to the

o-na [289]

Eybrbdbfbrjhebrbfjhfhrhrbrvhfhfjf

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3 years ago

What is the minimum number of degrees that a hexagon can be rotated so that it carried onto itself

Yuliya22 [10]

It has a rotational symmetry of 6

So minimum number of degrees = 360 / 6

= 60 degrees

4 0

3 years ago

Read 2 more answers

If you are receiving $57.94 in change, what is the lowest combination of bills and coins that will still give you the correct ch

mestny [16]

1 50 dollar bill

1 5 dollar bill

2 1 dollar bills

3 quarters

1 dime

1 nickel

4 pennies

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3 years ago

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti

jeyben [28]

Answer:

We need a sample of size at least 13.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$