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3 years ago

Solve the equation 17=5k-3

Mathematics

2 answers:

Gre4nikov [31]3 years ago

6 0

Answer:

k = 4

Step-by-step explanation:

17 = 5k - 3

+3. +3

20/5 = 5k/5

4 = k

k = 4

Mila [183]3 years ago

4 0

Answer: k = 4

17=5k-3

1. In order to solve this linear equation, we'll need to group all the variable terms on one side, and all the constant terms on the other side of the equation. In the problem, term -3 will be moved on the left side. Notice the sign changes when you move it to the left.

17=5k-3 ⇒ 17 + 3 = (5k - 3) + 3

20 = 5k - 3 + 3

2. Add the numerical terms.

20 = 5k - (3 + 3) = 6. Note: We don't need the six in the 3rd step.

20 = 5k

3. Switch the variable term. (Ex: 3x + 9 ⇒ 9 + 3x)

20 = 5k ⇒ 5k = 20

5k = 20

4. We will need to isolate the variable, to do that, get rid of the coefficient that it multiples. We can do that by dividing 5 by both sides.

5k/5 ÷ 20/5 = 1k/4

5k/5

5. We'll need to reduce 5k/5 to its lowest terms. We can divide the factors that appear both in the numerator and denominator. If we look at 5k/5, 5 appears both in the numerator and denominator.

5k/5 ⇒ k = 2²· 5/5

6. Lastly, we can do the same thing as we did in the 5th step. That'll lead us to k = 2²

7. Solve for 2²

2 x 2 = 4.

Hence, our answer is 4.

Got any questions? Tell me in the comments.

~ Lily, from Brainly.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Let p = proportion of households that have three cell phones NOT known.

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Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

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where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

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The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

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Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

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