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Valentin [98]
2 years ago
12

Calculate the volume of the composite shape please please help giving brainliest!!

Mathematics
1 answer:
Ilya [14]2 years ago
8 0

{\qquad\qquad\huge\underline{{\sf Answer}}}

There are two hemispheres in there with radius 3 cm, so we can consider it as a whole one. and the other shape in between is Cylinder with radius 3 cm and height 4 cm.

Volume of whole ahape = volume of Cylinder + volume of sphere ~

<u>Volume</u> <u>of</u> <u>sphere</u> :

\qquad \sf  \dashrightarrow \:  \cfrac{4}{3} \pi {r}^{3}

\qquad \sf  \dashrightarrow \:  \cfrac{4}{3}  \sdot3.14 \sdot {(3)}^{3}

\qquad \sf  \dashrightarrow \: (4 )\sdot(3.14) \sdot(9)

\qquad \sf  \dashrightarrow \: 113.04 \:  \: cm {}^{3}

<u>Volume</u> <u>of</u> <u>Cylinder</u> :

\qquad \sf  \dashrightarrow \: \pi {r}^{2} h

\qquad \sf  \dashrightarrow \: 3.14 \cdot(3) {}^{2}  \sdot4

\qquad \sf  \dashrightarrow \: 3.14 \sdot(9) \sdot(4)

\qquad \sf  \dashrightarrow \: 113.04 \:  \: cm {}^{3}

<u>Volume</u> <u>of</u> <u>whole</u> <u>shape</u> :

\qquad \sf  \dashrightarrow \: 113.04 + 113.04 = 226.08 \:  \: cm {}^{3}

\qquad \sf  \dashrightarrow \:  226.08 \:  \: cm {}^{3}

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the table shows the number of yards a football player runs in each quater of a game find the mean number of yards the player run
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A simple random sample of 60 households in city 1 is taken. In the sample, there are 45 households that decorate their houses wi
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Answer:

The calculated value of z = - 0.197  falls in the critical region therefore we reject the null hypothesis and conclude that  at  the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays

against Ha : p1≠ p2  (claim)  ( two sided)

The significance level is set at ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

or Z∝= 0.05/2= ± 1.96

The test statistic is

Z = p1-p2/√pq(1/n1 +1/n2)

p1= proportions of households  decorating in city 1  = 45/60=0.75

p2= proportions of households  decorating in city 2 = 40/50= 0.8

p = the common proportion on the assumption that the two proportion are same.

p =   \frac{n_1p_1 +n_2p_2}{n_1+n_2}

Calculating

p =60 (0.75) + 50 (0.8) / 110

p=  45+ 40/110= 85/110 = 0.772

so  q = 1-p= 1- 0.772= 0.227

Putting the values in the test statistic and calculating

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z= -0.05/√ 0.175244 ( 110/300)

z= -0.05/0.25348

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