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4 years ago

What is this problem17-3(w+5)=6(w+5)-2w Answer

Mathematics

1 answer:

Basile [38]4 years ago

6 0

Answer:

w = -4

Step-by-step explanation:

Solve for w:

17 - 3 (w + 5) = 6 (w + 5) - 2 w

-3 (w + 5) = -3 w - 15:

-3 w - 15 + 17 = 6 (w + 5) - 2 w

Grouping like terms, -3 w - 15 + 17 = (17 - 15) - 3 w:

(17 - 15) - 3 w = 6 (w + 5) - 2 w

17 - 15 = 2:

2 - 3 w = 6 (w + 5) - 2 w

6 (w + 5) = 6 w + 30:

2 - 3 w = 6 w + 30 - 2 w

Grouping like terms, 6 w - 2 w + 30 = (6 w - 2 w) + 30:

2 - 3 w = (6 w - 2 w) + 30

6 w - 2 w = 4 w:

2 - 3 w = 4 w + 30

Subtract 4 w from both sides:

2 + (-3 w - 4 w) = (4 w - 4 w) + 30

-3 w - 4 w = -7 w:

-7 w + 2 = (4 w - 4 w) + 30

4 w - 4 w = 0:

2 - 7 w = 30

Subtract 2 from both sides:

(2 - 2) - 7 w = 30 - 2

2 - 2 = 0:

-7 w = 30 - 2

30 - 2 = 28:

-7 w = 28

Divide both sides of -7 w = 28 by -7:

(-7 w)/(-7) = 28/(-7)

(-7)/(-7) = 1:

w = 28/(-7)

The gcd of 28 and -7 is 7, so 28/(-7) = (7×4)/(7 (-1)) = 7/7×4/(-1) = 4/(-1):

w = 4/(-1)

Multiply numerator and denominator of 4/(-1) by -1:

Answer: w = -4

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Jennifer spends 1/4 of the day at school then 3/4 with her friends how much time does she do in a day

Basile [38]

She spends 1/4 of her day at school, this would be 8 hours out of 24 hours. which would be 8/24=1/4. therefore she spends 14 hours with her friends (3/4)

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3 years ago

In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence

Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - α)% confidence interval for population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$