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3 years ago

Gary ate 41/50 of his candy bar. What percent of the candy bar did Gary eat? *

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

8 0

1) 82%
2) 1/5
3) 60%
4) 50%
5) 300
6) 18
7) 40
8) 15
9) 112.5
10) 96

ruslelena [56]3 years ago

3 0

1. 82%
2. 1/5
3. 60%
4. 50%
5. 6.75
6. 300
7. 18
8. 40
9. 60
10. 112.5
11. 96

Hope this helps!

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allochka39001 [22]

Answer:

Step-by-step explanation:

Perimeter of quarter circle = r + r + $\frac{1}{4}*2*\pi *r$

$= 10 + 10 + \frac{1}{4}*2*3.14*10\\\\= 10 + 10 + 3.14*5\\\\\\= 10 + 10 + 15.70\\\\= 35.70 cm$

2) d = 2*10 = 20 cm

Perimeter of semicircle = diameter + (1/2) * circumference of circle

$= 20 + \frac{1}{2}* 2*\pi *r\\\\= 20 + \frac{1}{2}*2*3.14*10\\\\= 20 + 3.14*10\\\\= 20 + 31.4\\\\= 51.40 cm$

3) Perimeter of three-quarter = r + r + (3/4)* circumference of circle

$= 10 + 10 + \frac{3}{4}*2*\pi *r\\\\= 10 + 10 + \frac{3}{4} * 2 * 3.14 * 10\\\\= 10 + 10 + 3 * 3.14 * 5\\\\= 10 + 10 + 47.1\\\\= 67.1 cm$

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3 years ago

MATH QUESTION DOWN BELOW

Gnom [1K]

8/9+(-5/6)= 8/9-5/6 = 1/18
Therefore, 1/18/(1/6)
1/18*1/6 (cancel out)
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3 years ago

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data

Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

The interval will be

$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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