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3 years ago

Julia charged $612 to her credit card. The cost of her next purchase caused her to go over her credit limit of $2000.

Mathematics

2 answers:

Fynjy0 [20]3 years ago

7 0

She has already spent 612, and by adding C 612 has gone ABOVE 2000
So 612 + C MUST be Greater Than 2000
The answer is D

gregori [183]3 years ago

7 0

Answer: Option D.

Step-by-step explanation:

So if she charged $612 to the card, this means that she spent $612 with her credit card.

Now, she purchases something with a cost of c dollars, and she goes over the limit of $2000

So we must have that:

$612 + c is greater than $2000

this can be written as:

$612 + c > $2000

the correct option is option D.

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Show how to solve the problem 378×6 Using Pl. value with regrouping explain how you knew when to regroup

Fynjy0 [20]

378
x 6
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I think that's how i learned it

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Answer:

it is a :) hope this helped :)

Step-by-step explanation:

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2 years ago

A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is current

lapo4ka [179]

Using the z-distribution, we have that:

The hypothesis test is:

$H_0: \mu = 7.4$

$H_1: \mu < 7.4$

The p-value is of 0.0668.

The critical value is $z^{\ast} = -1.28$

b) Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is tested if the mean is of 7.4 minutes, that is:

$H_0: \mu = 7.4$

At the alternative hypothesis, it is tested if the mean is lower than 7.4 minutes, that is:

$H_1: \mu < 7.4$

We have the standard deviation for the population, hence, the z-distribution is used.

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the sample.

n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36$ .

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}$

$z = -1.5$

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a left-tailed test, as we are testing if the mean is less than a value, the critical value with a significance level of 0.1 is of $z^{\ast} = -1.28$ .

Item b:

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

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1 year ago

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lesantik [10]

Answer:

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Step-by-step explanation:

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2 years ago

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≥-5

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2 years ago