Martin used a coordinate plane to design the garden shown. Each unit on the grid represents 10 yards.

Geometry help please! Match the reasons with the statements.

s344n2d4d5 [400]

Answer:

you need to put the reasons and statements

6 0

3 years ago

PLEASE HELP ASAAAP!!!!!!!q

Studentka2010 [4]

Answer:

A:a , s:soon ,A : as,P: possible

6 0

3 years ago

Using no other digits except 2 and 5 write down, if possible, a three-digit multiple of 6

Komok [63]

Answer: 255

Step-by-step explanation:

6 0

3 years ago

Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio

Kryger [21]

Answer:

From the given data, there is not enough evidence to prove that there is a statistically significant difference between the two population means

Step-by-step explanation:

The number of younger moms, in the study = 837

The average weight gain of younger moms, $\overline x$ ₁ = 30.67 pounds

The standard deviation of the weight gain of younger moms, s₁ = 14.69 pounds

(30.67 - 28.52)/√((14.69^2)/837 + 13²/143))

The number of younger moms, in the study = 143

The average weight gain of mature moms, $\overline x$ ₂ = 28.52 pounds

The standard deviation of the weight gain of mature moms, s₂ = 13 pounds

The test statistic for the difference in two populations is given as follows;

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_1^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Therefore, we get;

$t=\dfrac{(30.67-28.52)}{\sqrt{\dfrac{14.69^{2} }{837}-\dfrac{13^{2}}{143}}} \approx 1.7919$

The test statistic ≈ (1.7919)

Using a graphing calculator, we get;

The critical-t = ±1.971379, p = 0.07459697

Therefore, given that the test statistic, (1.7919), < critical-t (0.07459697), we fail to reject the null hypothesis, therefor, the given data does not provide convincing evidence that there is a significant difference between the two population means

7 0

3 years ago

henry has 68 miles to destination after 45 minutes and 51.5 miles to destination after 71 minutes of driving. How many miles to

Airida [17]

Let d represent the distance of the destination from the starting point.

After 45 min, Henry has already driven d-68 miles. After 71 min., he has already driven d-51.5 miles.

So we have 2 points on a straight line:

(45,d-68) and (71,d-51.5). Let's find the slope of the line thru these 2 points:

d-51.5 - (d-68) 16.5 miles
slope of line = m = ----------------------- = ------------------
71 - 45 26 min

Thus, the slope, m, is m = 0.635 miles/min

The distance to his destination would be d - (0.635 miles/min)(79 min), or

d - 50.135 miles. We don't know how far his destination is from his starting point, so represent that by "d."

After 45 minutes: Henry has d - 68 miles to go;

After 71 minutes, he has d - 51.5 miles to go; and

After 79 minutes, he has d - x miles to go. We need to find x.

Actually, much of this is unnecessary. Assuming that Henry's speed is 0.635 miles/ min, and knowing that there are 8 minutes between 71 and 79 minutes, we can figure that the distance traveled during those 8 minutes is

(0.635 miles/min)(8 min) = 5.08 miles. Subtracting thix from 51.5 miles, we conclude that after 79 minutes, Henry has (51.5-5.08), or 46.42, miles left before he reaches his destination.

7 0

3 years ago