A football team loses 5 yards on one play and then loses 8 yards on the nezt play. Write an addition expression that represents the change in the position of the team for the two plays. Then find the sum

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Answer 1

Answer: -5+-8+-13 yards in would be

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What’s the answer to this

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Answer:

Step-by-step explanation:

Given f(x) the f(x) + c represents a vertical translation of f(x)

• If c > 0 then a shift up of c units

• If c < 0 then a shift down of c units

Here c = - 6, thus a shift down, thus

g(x) = x² - 6 → A

I NEED HELP WITH THIS PLZZZ

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Answer:

The center of the circle is point: C=(1,5).

The radius of the circle is r=3.

Step-by-step explanation:

x² + y² -2x - 10y + 17 = 0

Find the center and radius of this circle

We have the equation

x² + y² -2x - 10y + 17 = 0

We collect like terms

x² - 2x + y² - 10y = -17

x² - 2x + (-2/2)² + y² - 10y + (-10/2)² = -17 + (-2/2)² + (-10/2)²

The center of the circle is point: C=(1,5).

The radius of the circle is r=3.

Simplify completely quantity 4 x squared minus 32 x plus 48 all over quantity 3 x squared minus 17 x minus 6.

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4x^2-32x+48
__________
3x^2-17x-6
⬇
4(x^2-8x+12)
__________
3x^2-17x-6

Answer:

$(4(x-2))/(3x+1)$

Step-by-step explanation:

i got it right on the test but I don't know the steps of solving it sorry

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would be defective. With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

Answers

Answer: 216

Step-by-step explanation:

The formula to find the sample size , if prior population proportion is known :-

$n=p(1-p)((z_(\alpha/2))/(E))^2$

Given : The prior proportion of defective parts : p= 0.10

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_(\alpha/2)=1.96$

Margin of error : $E=0.04$

Now, the required sample size will be :-

$n=0.1(0.9)(((1.96))/(0.04))^2=216.09\approx216$

Hence, the minimum required sample size = 216

Final answer:

To achieve a margin of error of .04 or less with a 95% confidence level when the defect rate is 10%, at least 217 samples need to be taken.

Explanation:

The question pertains to the field of statistics, specifically sample sizes and margin of error. In order to estimate the minimum sample size needed to achieve a desired margin of error of .04 or less, we can use the formula for sample size in proportions: n = (Z^2*p*(1-p))/E^2.

In this formula:

Z refers to the z-value which, for a 95% confidence level, is 1.96.
p is the estimated proportion of defective parts, which in this case is 0.1 or 10%.
E is the desired margin of error, which is 0.04.

Substitute the values into the formula: n = (1.96^2*0.1*0.9)/(0.04^2), yielding n=216.09.

Since we can't have a fraction of a sample, we round up to get n = 217. Therefore, we need a sample size of 217 to reach a margin of error of .04 or less with a 95% confidence level.