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pogonyaev
3 years ago
9

Tamara needs to buy motor oil to fill the 3 empty cylindrical barrels at her oil service center. Each barrel is 7 ft deep and ha

s a radius of 4 ft. What is the volume of oil needed? Use Test Image = 3.14.
Mathematics
1 answer:
aev [14]3 years ago
7 0
The total volume of the three cylinders is given by:
 V = (3) * (pi) * (r ^ 2) * (d)
 Where,
 r: cylinder radius
 d: cylinder depth
 Substituting values we have:
 V = (3) * (3.14) * (4 ^ 2) * (7)
 V = 1055.04 feet ^ 3
 Answer:
 the volume of oil needed is:
 V = 1055.04 feet ^ 3
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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
PLS HELP!!!!!!!!!!!!!!!!!! THIS IS DUE SOON AND I DONT GET IT
S_A_V [24]

Answer:

80

Step-by-step explanation:

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4 0
2 years ago
A bag contains 4 red marbles, 5 yellow marbles, and 6 blue marbles. three marbles are to be picked out randomly (without replace
gayaneshka [121]
A 50 50 percent , I think
6 0
3 years ago
Solve each equation.<br> 3a - 7 = 26<br> A=
Elanso [62]

Answer:

a = 11

Step-by-step explanation:

One is given the following equation:

3a - 7 = 26

Use inverse operations to solve this equation:

3a - 7 = 26

(add (7) to both sides)

3a = 26 + 7

Simplify,

3a = 26 + 7

3a = 33

Inverse operations,

3a = 33

(divide both sides by (3))

a = 33 / 3

Simplify,

a = 11

4 0
3 years ago
For how many movies is the cost of the plans the same?
sesenic [268]

Answer:

For 36 movies the cost of both the plans is same.

Step-by-step explanation:

Let us assume foe m movies, both the plans cost same.

Now, PLAN A:

Annual Fee = $45

Cost per movie  =  $2.50

⇒The cost of watching m movies  = m x (Cost of 1 movie)

= m x ($2.50) = 2.5  m

So, the total cost of Plan  A  = Annual Fee + Cost of m moves

                                                 = 45 + 2.50 m

PLAN B:

Cost per movie  =  $3.75

⇒The cost of watching m movies  = m x (Cost of 1 movie)

= m x ($3.75) = 3.75  m

ACCORDING TO QUESTION:

for m movies,  Cost of plan A = Cost of plan B

⇒45 + 2.50 m  = 3.75  m

or, 3.75 m  - 2.5 m = 45

or, m  = 45/1.25 = 36

or, m  = 36

Hence, for 36 movies the  cost of  both the plans is same.

5 0
3 years ago
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