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lana [24]
3 years ago
12

Solve Each Equation PLS SHOW WORK |5k-8| / 10 = 2 |5k-8| / 10 is a fraction btw

Mathematics
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

The maximum value of k = 5 ³/₅ and the minimum value of k = -2 ²/₅

Step-by-step explanation:

Given;

\frac{|5k-8|}{10} = 2\\\\|5k-8| = 20

Simplifying further;

5k - 8 = 20 ------- (1)

5k - 8 = -20------(2)

Solving equation (1)

5k - 8 = 20

5k = 28

k = 28 /5 = 5 ³/₅

Solving equation (2)

5k - 8 = -20

5k = -20 + 8

5k = -12

k = -12 / 5  = -2 ²/₅

Thus, the maximum value of k = 5 ³/₅ and the minimum value of k = -2 ²/₅

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9 - 3 * 2 + 4

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Which is the simplified form of the expression? Question 10 options: 9n – 2 9n + 2 15n – 2 15n – 8
KATRIN_1 [288]

Answer:

Combine -15 + 2n first, then you get -13 n.

Then work from -13n to subtracting -9n, that is -4n. Now all you have left is -4n = 8. Divide both sides by -4n/-4 8/-4. The answer is n = -2.

Step-by-step explanation:

3 0
3 years ago
N experiment was conducted to test the effect of a new stain remover. Twenty stains were treated with stain remover A and stain
Klio2033 [76]

Answer:

For this case the statistic calculated is t= 1.37

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{A} represent the mean for the sample for A

\bar X_{B} represent the mean for the sample B

s_{A} represent the sample standard deviation for the sample A

s_{B}= represent the sample standard deviation for the sample B

n_{A}=20 sample size for the group poisoned

n_{B}=20 sample size for the group unpoisoned

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two means are equal or not , the system of hypothesis would be:

Null hypothesis:\mu_{A} - \mu_{B} =0

Alternative hypothesis:\mu_{A} - \mu_{B} \neq 0

For this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

For this case the statistic calculated is t= 1.37

Statistical decision

The significance level is 0.1 \alpha=0.1, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

df=n_{A}+n_{B}-2

The p value on this case is given

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0,1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

5 0
3 years ago
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