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3 years ago

What is the approximate circumference of a circle with a diameter of 300 feet

1 answer:

6 0

Circumference = 2πr

Since you know the diameter is 300, you divide it by 2 to find the radius. Therefore, the radius is 150.
Now plug it into the formula: C = 2(3.14)(150)
C = 942

Answer: Circumference is 942 feet

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X+3y=28 when x =28 find the value for y plz help

Lisa [10]

Answer:

y = 0

Step-by-step explanation:

Plug in 28 for x. Solve for the equation:

x + 3y = 28

(28) + 3y = 28

Isolate the variable, y. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS. First, subtract 28 from both sides:

28 (-28) + 3y = 28 (-28)

3y = 0

Isolate the variable, y. Divide 3 from both sides:

(3y)/3 = (0)/3

y = 0

y = 0 is your answer.

4 0

3 years ago

The radius of a circle is 5 miles. What is the angle measure of an arc bounding a sector with area 10 square miles?

Lana71 [14]

10÷((22/7)×5^2))
=0.127

6 0

3 years ago

The weight of laboratory grasshoppers follows a Normal distribution, with a mean of 90 grams and a standard deviation of 2 grams

JulsSmile [24]

Answer:

95.44% of the grasshoppers weigh between 86 grams and 94 grams.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 90 grams and a standard deviation of 2 grams.

This means that $\mu = 90, \sigma = 2$

What percentage of the grasshoppers weigh between 86 grams and 94 grams?

The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 86. So

X = 94

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

$Z = \frac{94 - 90}{2}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

X = 86

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

$Z = \frac{86 - 90}{2}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544*100% = 95.44%

95.44% of the grasshoppers weigh between 86 grams and 94 grams.

5 0

3 years ago

Write a trinomial which is degree 2, where the coefficient of the highest degree term is 7.

Nady [450]

Answer:

7x^2 - 2x - 3.

Step-by-step explanation:

Starts with 7x^2 then a term in x then a constant.

4 0

3 years ago

A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample

fiasKO [112]

Answer:

5.05% probability that no more than 34% had received such an email.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .