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3 years ago

A triangle has side lengths of (8d – 3) centimeters, (6d + 2) centimeters, and

Mathematics

1 answer:

alisha [4.7K]3 years ago

3 0

The perimeter of a triangle is the sum of its side lengths.

The perimeter of the triangle is: $14d + 7f +7$

The sides of the triangle are:

Side 1 = (8d - 3) cm

Side 2 = (6d + 2) cm

Side 3 = (7f + 8) cm

The perimeter (P) is calculated as:

P = Side 1 + Side 2 + Side 3

So, we have:

$P = (8d - 3) + (6d + 2) + (7f + 8)$

Remove brackets

$P = 8d - 3 + 6d + 2 + 7f + 8$

Collect like terms

$P = 8d + 6d + 7f - 3 + 2 + 8$

$P = 14d + 7f +7$

Hence, the perimeter of the triangle is: $14d + 7f +7$

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The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum

Vadim26 [7]

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762$

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

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

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

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

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

X = 25

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

$Z = \frac{25 - 26}{0.762}$

$Z = -1.31$

$Z = -1.31$ has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

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

$-1.645 = \frac{X - 26}{0.762}$

$X - 26 = -1.645*0.762$

$X = 24.75$

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

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

$1.645 = \frac{X - 26}{0.762}$

$X - 26 = 1.645*0.762$

$X = 27.25$

Between $24.75 and $27.25.

3 0

3 years ago

If n(A) = P and n(B)=q then n(AxB) is (a ) p (b ) q ( c ) p+q ( d ) pq

Butoxors [25]

By way of example, suppose A = {1, 2, 3} and B = {a, b, c}. Then the Cartesian product of A and B is

A × B = {{1, a}, {1, b}, {1, c}, {2, a}, {2, b}, {2, c}, {3, a}, {3, b}, {3, c}}

That is, each element in A gets a pairing with each element in B, and for each pairing you have n(A) choices for the first element and n(B) choices for the second element.

So if n(A) = p and n(B) = q, then n(A × B) = pq.

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3 years ago

Find the volume of a right circular cone that has a height of 14.3 ft and a base with a diameter of 17.2 ft. Round your answer t

saveliy_v [14]

Answer:

Volume of the given cone is 1107.6 cubic feet.

Step-by-step explanation:

Formula to get the volume of a right circular cone is,

V = $\frac{1}{3}\pi r^{2}h$