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mars1129 [50]
3 years ago
7

I need help with this please

Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

diameter = 24 yards

Step-by-step explanation:

C = π·d

75.36 = 3.14d

d = 75.36/3.14

d = 24

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Let f(x)=15/(1+4e)^(-0.2x)<br><br> What is the y-intercept of the graph of f(x) ?
sergejj [24]
That means when x=0
e^-0.2x =e^0=1
So f(0)=15/5=3
5 0
4 years ago
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Need help with this question ​
Hatshy [7]

Answer:

1. He added 5 instead of subtracted 5.

2. He should have subtracted 5.

3. 2 2/3

Step-by-step explanation:

3x+5=13

subtract 5 to both sides 3x=8

divide both sides by 3 x=8/3 or 2 2/3

8 0
3 years ago
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. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
How do you do this question?
daser333 [38]

Answer:

B. 1/2

Step-by-step explanation:

\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}

If we plug in 0 for z, we get 0/0.  Apply l'Hopital's rule.

\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}

Now when we plug in 0 for z, we get:

\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}

4 0
3 years ago
What is the degree of 12x^4-8x+4x^2-3
lord [1]
Dat is 4th degree because x is raised to the 4th power and that is the highest power of the placholder

4th degree
6 0
4 years ago
Read 2 more answers
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