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Andru [333]
3 years ago
5

Which value is equivalent to V-25?A. -5iB. -5C. 5OD. 5i​

Mathematics
1 answer:
AlexFokin [52]3 years ago
6 0

Answer:

\sqrt{-25} = 5i

Step-by-step explanation:

Given

\sqrt{-25

Required

An equivalent

We have:

\sqrt{-25

Express -25 as -1 * 25

\sqrt{-25} = \sqrt{-1 * 25}

Split

\sqrt{-25} = \sqrt{-1} * \sqrt{25}

\sqrt{25} = 5, so:

\sqrt{-25} = \sqrt{-1} * 5

\sqrt{-1} = i --- complex numbers

So:

\sqrt{-25} = i * 5

\sqrt{-25} = 5i

<em>Hence, the equivalent is (D)</em>

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-(4  1/5)
5(4)+1 = 21
Thus the answer would be -21/5
4 0
3 years ago
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Which of these expressions are equivalent to 4x+2(2x−5)−(3−5x) ? Choose the TWO correct answers. 3x−13
Julli [10]

Answer: 4x+2(2x−5)−(3−5x) = 8x-10-3+5x

Step-by-step explanation: We need to find the equivalent expression to 4x+2(2x−5)−(3−5x).

Using distributive property as follows :

a(b+c) = ab+ac

4x+2(2x−5)−(3−5x) = 4x+4x-10-3+5x

= 8x-10-3+5x

Option (e)

Hence, the correct option is (e) "8x-10-3+5x"

3 0
3 years ago
The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees ba
blsea [12.9K]

Answer:

The test statistic is z = -2.11.

Step-by-step explanation:

Before finding the test statistic, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Group 1: Sample of 35, mean of 1276, standard deviation of 347.

This means that:

\mu_1 = 1276, s_1 = \frac{347}{\sqrt{35}} = 58.6537

Group 2: Sample of 35, mean of 1439, standard deviation of 298.

This means that:

\mu_2 = 1439, s_2 = \frac{298}{\sqrt{35}} = 50.3712

Test if there is a difference in productivity level.

At the null hypothesis, we test that there is no difference, that is, the subtraction is 0. So

H_0: \mu_1 - \mu_2 = 0

At the alternate hypothesis, we test that there is difference, that is, the subtraction is different of 0. So

H_1: \mu_1 - \mu_2 \neq 0

The test statistic is:

z = \frac{X - \mu}{s}

In which X is the sample mean, \mu is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that \mu = 0

From the two samples:

X = \mu_1 - \mu_2 = 1276 - 1439 = -163

s = \sqrt{s_1^2+s_2^2} = \sqrt{58.6537^2+50.3712^2} = 77.3144

Test statistic:

z = \frac{X - \mu}{s}

z = \frac{-163 - 0}{77.3144}

z = -2.11

The test statistic is z = -2.11.

7 0
3 years ago
1: What is Abby's total spent on transportation?
stealth61 [152]

Answer:

1. Total spent on transportation $222.2.

2. $ 192.5

3. She spends $30 over the guideline.

Step-by-step explanation:

As per the image attached:

Total monthly spent on transportation = Total monthly car payment + Spent on gas fuel

Total monthly spent on transportation = 120 + 37.9 + 28.4 + 35.9 = $222.2

So, Answer 1. Abby's total spent on transportation = $222.2

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\Rightarrow \dfrac{\text{Spent on transportation}}{\text{Total monthly income}}

\Rightarrow \dfrac{222.2}{1750}\times 100 =12.69\%

Calculating 11% as per the guideline:

1750 \times \dfrac{11}{100}\\\Rightarrow \$192.5

Answer 2. As per the guideline, she should spend $192.5 on transportation.

Difference between spending and guideline = $222.2 - $192.5 = $29.7

Answer 3. She spends $29.7 over the guideline.

8 0
3 years ago
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Answer:

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