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3 years ago

Anne has three more dollars than Freddy. Together they have $99. How many dollars does each have?

Mathematics

2 answers:

Tamiku [17]3 years ago

7 0

Anne has 51. Freddy has 48.

djyliett [7]3 years ago

6 0

Anne has 51 dollars and freddy has 48 dollars. You find the answer by doing, 99=x+(x+3)
96=2x
48=x
so you plug 48 into x

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Step-by-step explanation:

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3 years ago

Write a story for this graph. ( Use the outline plz)

Katyanochek1 [597]

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Step-by-step explanation:

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3 years ago

The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery tim

Hatshy [7]

Answer:

The expected cost is 152

Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

$f_Y(y ) = \frac{1}{5-1} = \frac{1}{4}$ if $1\leq y \leq 5$ and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

Recall that, by definition

$E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy$

Also, we are given that $C = 50+3Y+9Y^2$ . Recall the following properties of the expected value. If X,Y are random variables, then

$E(a+bX+cY) = a+bE(X)+cE(Y)$

Then, using this property we have that $E[C] = 50+3E[Y]+ 9E[Y^2]$ .

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

$E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3$

$E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}$

Then

$E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152$

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3 years ago

Assume that a hypothesis test to support the given claim will be conducted. Identify the type l error for the test. A skeptical

Mars2501 [29]

Answer:

Option D) Reject the claim that the proportion of Americans that have seen a UFO is equal to 5 in a thousand when that proportion is actually 5 in a thousand.

Step-by-step explanation:

We are given the following information:

A hypothesis test to support the given claim that the proportion of Americans that have seen a UFO is less than 5 in a thousand.

Type I error:

Type I error is the rejection of a true null hypothesis that is we reject the null hypothesis given it is true.
It is also known as a false positive.

First, we design the null and the alternate hypothesis

$H_{0}: \text{ The proportion of Americans that have seen a UFO is equal to 5 in a thousand}\\H_A: \text{ The proportion of Americans that have seen a UFO is less than 5 in a thousand}$

Thus, for the given hypothesis test, we describe type I error as:

Option D) Reject the claim that the proportion of Americans that have seen a UFO is equal to 5 in a thousand when that proportion is actually 5 in a thousand.

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4 years ago