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zhenek [66]
2 years ago
10

NEED HELP ASAP failing

Mathematics
1 answer:
fredd [130]2 years ago
5 0
Need to do the 2^3 first

answer is A
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A cone-shaped pile of gravel has a diameter of 30 m and a height of 9.1 m.
Arte-miy333 [17]
D/2=r
vcone=(1/3)hpir^2
given
d=30
h=9.1
d/2=30/2=15=r

v=(1/3)9.1pi15^2
v=(9.1/3)pi225
v=682.5pi
use 3.141592 to aprox pi
v=2144.13654
the closese is the first one
answer would be 2120 m³
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Darlene spent 20 minutes jogging and 10 minutes cycling and burned 400 calories . The next day she did 10 minutes of jogging and
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This is very spread out just to show properly but you could get it smaller haha

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Amina used the quadratic formula to solve an equation. Her result shows two solutions that are complex numbers with imaginary pa
Mama L [17]

A quadratic equation has either two different real roots, one real root, or two conjugate complex roots (this is the case when the discriminant is negative, i.e. when you have no real roots).

Two conjugate complex roots have the same real part and opposite imaginary parts. So, the solutions to Amina's equation will be in this form:

a+bi,\ \ a-bi

For some a,b\in\mathbb{R},\ b\neq 0

6 0
2 years ago
Determine the value of (tangent of 88 degrees, 22 minutes, and 45 seconds). In this case, minutes are 1/60 of a degree and secon
mariarad [96]

Answer:

  c.  35.34015106

Step-by-step explanation:

As with many problems of this nature, you only need to get close to be able to choose the correct answer. 22 minutes 45 seconds is just slightly less than 1/2 degree (30 minutes), so the tangent value will be just slightly less than tan(88.5°) ≈ 38. The appropriate choice is 35.34015106.

If you need confirmation, you can find tan(88°) ≈ 29, so you know the answer will be between 29 and 38.

__

The above has to do with strategies for choosing answers on multiple-choice problems. Below, we will work the problem.

The angle is (in degrees) ...

  88 + 22/60 +45/3600 = 88 + (22·60 +45)/3600 = 88 +1365/3600

  ≈ 88.3791666... (repeating) . . . . degrees

A calculator tells you the tangent of that is ...

  tan(88.3791666...°) ≈ 35.3401510614

Many calculators will round that to 10 digits, as in the answer above. Others can give a value correct to 32 digits. Spreadsheet values will often be correct to 15 or 16 digits.

4 0
2 years ago
The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi
SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so n = 1210.

The mean of the sample is 1030.

The probability of a success is \pi = \frac{1030}{1210} = 0.8512.

The standard deviation of the sample is s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when X = 1030.

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

\mu = 0.82(1210) = 992.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{1030 - 992.2}{12.38}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989.This means that the county result is unusually high.

4 0
2 years ago
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