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2 years ago

Solve please will mark brainliest

Mathematics

1 answer:

Whitepunk [10]2 years ago

6 0

Hello! for this equation the formula is:
a times (a+b)= c times (c+d)

4 times (4+18)=4 times 22=88
88=6 times (6+x)=6x+36
88=6x+36
88-36=6x
6x=52

x= 8 and 2/3

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the figure below shows two half-circles at the ends of a rectangle with the dimensions shown. which is the closet to the area of

Pani-rosa [81]

You get 92.56 <span>You add the area of the rectangle to the area of a circle

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8 0

3 years ago

2 ratios equivalent to 8\3

borishaifa [10]

Some equivalent fractions of 8/3 are:
8/3 = 16/6 = 24/9 = 32/12 = 40/15 = 48/18 = 56/21 = 64/24 = 72/27 = 80/30 = 88/33 = 96/36 = 104/39 = 112/42 = 120/45 = 128/48 = 136/51 = 144/54 = 152/57 = 160/60

3 0

3 years ago

Someone help, the question is on the attachment.

gavmur [86]

The answer is y = 2.

6 0

3 years ago

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

Help plz ? Enter the slope of the line

Dmitrij [34]

(-2,4), (2,5)

Slope equation : (y₂-y₁) / (x₂-x₁)

Plug in the values.

m = slope

m = (5-4) / (2 + 2)

Simplify.

m = 1/4

So, your slope is 1/4

~Hope I helped!~

4 0

3 years ago