Please help me with this algebra 2 question

Which equation, when solved, results in a different value of x than the other three? O -7-9-20 0 + 5 3 7 X=-20 4 O 3 4. - (b)*--

Molodets [167]

Answer:

1

Step-by-step explanation:

6 0

3 years ago

HELP PLEASEE NOWWWWEWEWNENSH PLEASEE OMG

Irina-Kira [14]

Answer: None.

Step-by-step explanation: It is translated up one unit and then reflected across the y axis

4 0

3 years ago

The elephant eats 220 pounds of food a day costing $45. The tax on the food is 12%. How much does the zoo spend on elephant food

OLEGan [10]

So 220 pounds of food a day is 45$ a day so the tax is 12% so since 45$ a day and 12% for tax so if you minus it is 5.4

6 0

3 years ago

What is the relationship of the line to the line 6x-2y=7 ? Are they parallel, perpendicular or neither

Mila [183]

Answer:

you only put one line bro

8 0

3 years ago

In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hir

Ede4ka [16]

Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}= p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

$\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204$

So, the sampling distribution of sample proportion is $\hat p\sim N(0.75,0.0204^{2})$ .

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

$P(p-0.04$

$=P(-1.96$

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

$\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306$

So, the sampling distribution of sample proportion is $\hat p\sim N(0.75,0.0306^{2})$ .

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

$P(p-0.04$

$=P(-1.31$

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

6 0

3 years ago