15 units right of 0 what is the answer

Witch has a greater absolute value 33 dollars and -52 dollars

lakkis [162]

Answer:

-52

Step-by-step explanation:

the absolute value is always positive and so 33 would remain 33, but -52 would change to 52 which is greater than 33.

6 0

3 years ago

Read 2 more answers

Solve the quadratic function by completing the square. what are the missing pieces in the steps? –32 = 2(x2 10x) –32 = 2(x2 10x

NemiM [27]

An equation is formed of two equal expressions. The missing pieces of the solution of the quadratic equation are 50, 3, and -8.

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

To fill the missing piece of the solution of the quadratic equation, you need to balance each of the equations. Therefore, the solution to the problem can be written as,

$-32 = 2(x^2 + 10x)\\\\-32 + 50 = 2(x^2 + 10x + 25)\\\\18 = 2(x + 5)^2\\\\9 = (x + 5)2\\\\\pm 3= x + 5\\\\x = -2\ \ {\rm or}\ \ x = -8$

Thus, the solution of the quadratic equation is -2 or -8.

Hence, the missing pieces of the solution of the quadratic equation are 50, 3, and -8.

Learn more about Equation:

brainly.com/question/2263981

6 0

2 years ago

A cylindrical flower vase has a height of 8 inches and a radius of 2 inches what is the volume of the vase?

natta225 [31]

Answer:

1008.48 in ^2

Step-by-step explanation:

Given

Its a cylindrical vase .
Radius = 2 inches
height = 8 inches .

Formula

$\bold { \purple {volume = \pi \: {r}^{2}h \: }}$

Lets assume pi to be 3.14

$v = 3.14 \times {2}^{2} \times 8 \\ \\ \\ v = 3.14 \times 4 \times 8 \\ \\ \\ v = 3.14 \times 32 \\ \\ \\ v = 1008.48 {in}^{2}$

6 0

3 years ago

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given sta

seropon [69]

Answer:

The margin of error is of 0.0184 = 1.84%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

In this question:

$\pi = \frac{1672}{2388} = 0.7, n = 2388$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{0.7*0.3}{2388}}$

$M = 0.0184$

The margin of error is of 0.0184 = 1.84%.

6 0

3 years ago

A sample of 900900 computer chips revealed that 66f% of the chips fail in the first 10001000 hours of their use. The company's p

Alik [6]

Answer:

No, there is not enough evidence at the 0.02 level to support the manager's claim.

Step-by-step explanation:

We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

Null Hypothesis, $H_0$ : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}

Alternate Hypothesis, $H_1$ : p 0.68 {means that the actual percentage that fail is different from the stated percentage}

The test statistics we will use here is;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, p = actual percentage of chip fail = 0.68

$\hat p$ = percentage of chip failed in a sample of 900 chips = 0.66

n = sample size = 900

So, Test statistics = $\frac{0.66 -0.68}{\sqrt{\frac{0.66(1- 0.66)}{900} } }$

= -1.267

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.

Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.

5 0

4 years ago