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SVETLANKA909090 [29]
2 years ago
12

7x67 3x57 6x 6 4 x0 15X4

Mathematics
2 answers:
Ad libitum [116K]2 years ago
7 0

Answer:

7x67 = 469

3x57 = 171

6x6   = 36

4x0   = 0

15x 4 = 60

Step-by-step explanation:

8_murik_8 [283]2 years ago
4 0

Answer:

1. 469 2. 171 3. 36 4. 0 5. 60

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Val needs to find the area enclosed by the figure. The figure is made by attaching semicircles to each side of a 42-m-42-m squar
oee [108]

The area enclosed by the figure is 4533.48 square meters.

<u>Step-by-step explanation:</u>

Side length of the square = 42m

The semicircle is attached to each side of the square. So the diameter of the semicircle is the length of the square.

Radius of the semicircle = 21m

Area of the square = 42 x 42 = 1764 square meters

Area of 1 semicircle = π(21 x 21) /2

= (3.14) (441) /2

= 1384.74/2

= 692.37 square meters

Area of 4 semicircle = 4 x 692.37

= 2769.48 square meters

Total area = 1764 + 2769.48

= 4533.48 square meters

The area enclosed by the figure is 4533.48 square meters.

7 0
2 years ago
$10 + ($6-n) if n =$3<br> What is the final product <br> Of this mathematical question?
kap26 [50]
If n = $3 then plug it in
$10 + ($6-$3)
$6-$3 = $3
$10 + $3 = $13
5 0
3 years ago
Read 2 more answers
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
The average score on a standardized test is 750 points with a standard deviation of 50 points. What is the probability that a st
nalin [4]
The probability that student scored more than 850 we shall proceed as follows:
z=(x-μ)/σ
where:
x=850
μ=750
σ=50
thus
z=(850-750)/50
z=2
thus
P(x>850)=1-P(x<850)=1-P(z<2)=1-0.9772=0.0228

Answer: P(x>850)=0.0228
4 0
2 years ago
Read 2 more answers
Please help me answer this ty. ​
rodikova [14]

please mark me as brainlest

7 0
3 years ago
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