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3 years ago

What is the average of these four dollar amounts to the nearest cent? $400 $323.69 $7,352 $19.99

Mathematics

2 answers:

agasfer [191]3 years ago

7 0

Average=$(400+323.69+7352+19.99)÷4
=$2023.92
≈$2023.9

iogann1982 [59]3 years ago

5 0

$2,023.92 would be the numerical conclusion when dealing in terms of money. If it's in values and needed rounding, it would be 2,023.9.

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What two numbers add up to 10 and multiply to 21

WITCHER [35]

Answer:The answer is 3 & 7

Step-by-step explanation:3 x 7 = 21

3 + (7) = 10

Are you asking because you are trying to figure out how to factor the following quadratic equation?

x2 + 10x + 21 = 0

If so, the solution to factor the quadratic equation above is:

(X + 3 ) (X + 7)

To summarize, since 3 and 7 multiply to 21 and add up 10, you know that the following is true:

x2 + 10x + 21 = (X + 3 ) (X + 7)

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3 years ago

Suppose you are given two quadrilaterals, ABCD EFGH. If the similarity ratio of ABCD to EFGH is 4:3, and if EH = 60, find AD.

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3 years ago

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Step-by-step explanation:

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3 years ago

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Klio2033 [76]

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3 years ago

A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that

mafiozo [28]

Answer:

(a) The significance level of the test is 0.002.

(b) The power of the test is 0.3487.

Step-by-step explanation:

We are given that a coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that the probability is not 0.5.

The test rejects the null hypothesis if either 0 or 10 heads are observed.

Let p = probability of obtaining head.

So, Null Hypothesis, $H_0$ : p = 0.5

Alternate Hypothesis, $H_A$ : p $\neq$ 0.5

(a) The significance level of the test which is represented by $\alpha$ is the probability of Type I error.

Type I error states the probability of rejecting the null hypothesis given the fact that the null hypothesis is true.

Here, the probability of rejecting the null hypothesis means we obtain the probability of observing either 0 or 10 heads, that is;

P(Type I error) = $\alpha$

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

Also, the event of obtaining heads when a coin is thrown 10 times can be considered as a binomial experiment.

So, X ~ Binom(n = 10, p = 0.5)

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

$\binom{10}{0}\times 0.5^{0} \times (1-0.5)^{10-0} +\binom{10}{10}\times 0.5^{10} \times (1-0.5)^{10-10}$ = $\alpha$

$(1\times 1\times 0.5^{10}) +(1 \times 0.5^{10} \times 0.5^{0})$ = $\alpha$

$\alpha$ = 0.0019

So, the significance level of the test is 0.002.

(b) It is stated that the probability of heads is 0.1, and we have to find the power of the test.

Here the Type II error is used which states the probability of accepting the null hypothesis given the fact that the null hypothesis is false.

Also, the power of the test is represented by (1 - $\beta$ ).

So, here, X ~ Binom(n = 10, p = 0.1)

$1-\beta$ = P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true)

$1-\beta$ = $\binom{10}{0}\times 0.1^{0} \times (1-0.1)^{10-0} +\binom{10}{10}\times 0.1^{10} \times (1-0.1)^{10-10}$

$1-\beta$ = $(1\times 1\times 0.9^{10}) +(1 \times 0.1^{10} \times 0.9^{0})$

$1-\beta$ = 0.3487

Hence, the power of the test is 0.3487.

3 0

3 years ago