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N76 [4]
3 years ago
13

Please answer i will put you the brainliast

Mathematics
1 answer:
adell [148]3 years ago
7 0
2/3x2/3=4/9 (0.44)decimal answer
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Six friends went to a baseball game.The price of the admission per person was $x.Three of the friends bought a hot dog for $3. W
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C(cost)=6x+9
9 is the cost of three hot dogs
6x represents 6 tickets and their cost
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Equivalent fractions to 28/32
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7/8 there you go the right answer xD
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In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study
DIA [1.3K]

Answer:

z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

C. between 0.01 and 0.05.

Step-by-step explanation:

Information provided

X_{1}=1642 represent the number of smokers from the sample in 1995

X_{2}=1415 represent the number of smokers from the sample in 2010

n_{1}=4276 sample from 1995

n_{2}=3908 sample from 2010  

p_{1}=\frac{1642}{4276}=0.384 represent the proportion of smokers from the sample in 1995

p_{2}=\frac{1415}{3908}=0.362 represent the proportion of smokers from the sample in 2010

\hat p represent the pooled estimate of p

z would represent the statistic    

p_v represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

The statistic is given by:

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374  

Replacing the info given we got:

z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

C. between 0.01 and 0.05.

7 0
3 years ago
In a shipment of 22 smartphones, 2 are defective. How many ways can a quality control inspector randomly test 4 smartphones, of
Eddi Din [679]

Answer:

Therefore the required ways are =190

Step-by-step explanation:

Combination: Combination is the number of selection of items from a collection of items where the order of selection does not matter.

Total number of smartphones =22

Defective phone = 2

Non defective phone = (22-2) =20

The control inspector randomly test 4 smartphones of which 2 are defective.

Non defective phone = 2

The ways to select 2 non defective phone is

^{20}C_2=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!}=\frac{19\times 20}{2} =190

The ways to select 2 defective phone is= ^2C_2=1

Therefore the required ways are = (190×1) =190

5 0
4 years ago
8n+?=0<br> plz solve i will mark brainliest
Ksivusya [100]

Answer:

negative 8

Step-by-step explanation:

7 0
3 years ago
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