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3 years ago

10

You have a gift hard that is $20 and you want to buy a sweater that is 18.74 with 6% tax do you have enough PLEASE SHOW WORK

1 answer:

bekas [8.4K]3 years ago

3 0

To give you the total cost of the sweater you do:

18.74 × 1.06 = 19.8664

Which will be rounded to $19.87 because the measurement is currency (2 d.p.).

Therefore, you do have enough because $20 > $19.87.

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To make fuel for the main engines of a space shuttle, 102,619.377 kilograms of liquid hydrogen and 616,496.4409 kilograms of liq

aleksandrvk [35]

Answer:

There are 719,115.8179kg of fuel stored in the external tank

Step-by-step explanation:

102619.377 + 616496.4409 = 719,115.8179

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3 years ago

Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give t

Lapatulllka [165]

Answer:

$P(\bar X >80)=P(Z>2.143)=1-P(z$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

$X \sim N(\mu=74, \sigma=14)$

The sampling distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

The deduction is explained below we have this:

$E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu$

$Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)$

Since the variance for each individual observation is $Var(x_i)=\sigma^2$ then:

$Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}$

And then for this special case:

$\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)$

We are interested on this probability:

$P(\bar X >80)$

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

Applying this we have the following result:

$P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)$

And using the normal standard distribution, Excel or a calculator we find this:

$P(Z>2.143)=1-P(z$

6 0

4 years ago

One minus the quotient of one and a number x

Arada [10]

1 - (1 divided by 3)

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4 years ago

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Fifty students were polled on their favorite after-school activity. The results are in the table. What is the probability that a

Kay [80]

Answer:

4/13

Step-by-step explanation:

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3 years ago

Read 2 more answers

-2(v - 2) = -3 – 2v

liq [111]

Answer:

no solution

Step-by-step explanation:

Given

- 2(v - 2) = - 3 - 2v ← distribute left side

- 2v + 4 = - 3 - 2v ( subtract 4 from both sides )

- 2v = - 7 - 2v ( add 2v to both sides )

0 = - 7 ← not possible

This indicates the equation has no solution

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3 years ago