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3 years ago

What property or properties are not used when solving the following equation -3(4x-8)=-36

Mathematics

2 answers:

Tasya [4]3 years ago

8 0

You’re answer would be D. Distributive property ! Hope this helps have a great day .

BigorU [14]3 years ago

6 0

Either B or D I’m not really sure but I know it’s one of this answers

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To decrease the impact on the environment, factory chimneys must be high enough to allow pollutants to dissipate over a larger a

Komok [63]

Answer:

The probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

Step-by-step explanation:

Let the random variable X be defined as the height of chimneys in factories.

The mean height is, μ = 100 meters.

The standard deviation of heights is, σ = 12 meters.

It is provided that a random sample of n = 40 chimney heights is obtained.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

Since the sample selected is quite large, i.e. n = 40 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean heights of chimneys.

$\bar X\sim N(\mu_{\bar x},\ \sigma^{2}_{\bar x})$

Compute the probability hat the sample mean height for the 40 chimneys is greater than 102 meters as follows:

$P(\bar X>102)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}})>\frac{102-100}{12/\sqrt{40}})$

$=P(Z>1.05)\\=1-P(Z$

*Use a z-table fr the probability.

Thus, the probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

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3 years ago

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laila [671]

Answer: This is rly confusing

Step-by-step explanation:

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3 years ago

Evaluate the expression for x=5, y=-3 and z=7 7 - 3y – 3z

andrew11 [14]

Answer:

Step-by-step explanation:

7 - 3(-3) - 3(7) = 7 + 9 - 21 = 16 - 21 = -5

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3 years ago

Read 2 more answers

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$