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UkoKoshka [18]
4 years ago
11

What is the area of sector GPH?

Mathematics
1 answer:
jekas [21]4 years ago
7 0
The area of the entire circle = pi r^2
The area of the shaded area = (40/360) * pi r^2
r = 9 cm

Area of the shaded area = 1/9 * 3.14 * 9^2
Area of the shaded area = 3.14 * 9
Area of the shaded area = 28.26

Notice one of the 9s disappeared. Where did it go?
You could begin the problem by writing out 9^2
Area = 1/9 * 3.14 * 9 * 9 Notice that 1/9 will cancel out 1 of the nines.

28.26 yds = Shaded area. 
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Solve for p.<br> –<br> 19p–2p+16p+12=<br> –<br> 18<br> p=
fomenos

Answer:

6

<em>BRAINLIEST, PLEASE!</em>

Step-by-step explanation:

-19p - 2p + 16p + 12 = -18

-5p + 12 = -18

-5p = -30

p = 6

7 0
3 years ago
Read 2 more answers
How to do 2×+7+2×=19
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Apply the product rule to
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Tap for more steps...
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Apply the product rule to
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The result can be shown in multiple forms.
Exact Form:
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Decimal Form:
0.36253492
…
0.36253492
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3 0
3 years ago
Read 2 more answers
Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s
stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

8 0
3 years ago
A pizza parlor sold 38 pizzas during a dinner hour. If each pizza contained 8 slices, how many slices of pizza were sold?
AlladinOne [14]
38 x 8 = 304
So the answer is b
4 0
3 years ago
Read 2 more answers
if the lines shown below are perpendicular if the green line has a slope of -1/4 what is the slipped of the red line
insens350 [35]

Answer:

C; 4

Step-by-step explanation:

Mathematically, when two lines are perpendicular, the product of their slopes equal -1

in that case, we have it that;

m1 * m2 = -1

where m1 is the slope of the first line and m2 is the slope of the second line

Hence,

m2* -1/4 = -1

-m2/4 = -1

-m2 = -4

m2 = 4

7 0
3 years ago
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