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3 years ago

PLEASE HELP!! THIS IS MY LAST QUESTION!! WILL GIVE BRAIN!!

Mathematics

1 answer:

Brums [2.3K]3 years ago

7 0

Answer:

its answer 1.

Step-by-step explanation:

please mark me brainliest thank you for your time and have a nice rest of your day :)

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What is the following sum? 5(3√x)+9(3√x)

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5(3√x)+9(3√x)
=3√x(5+9)
=14(3√x)

answer is C. third choice 14(3√x)

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2 years ago

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Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come f

AlladinOne [14]

Answer:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

$\hat p=0.64$ estimated proportion of interest

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion is higher than 0.5:

Null hypothesis: $p =0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

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3 years ago

Suppose that a box contains 6 cameras and that 3 of them are defective. A sample of 2 cameras is selected at random. Define the

mezya [45]

Answer: 1/5, 1/2, 0.

Step-by-step explanation:

given data:

no of cameras = 6

no of cameras defective = 3

no of cameras selected = 2

Let p(t):=P(X=t)

p(2)=m/n,

m=binomial(3,2)=3!/2!= 3

n=binomial(6,2)=6!/2!/4! = 15

p(3)= 3/15

= 1/5.

p(1)=m/n,

m=binomial(6,1)*binomial(2,2)=6!/1!/4!*2!/2!/0!= 7.5

n=binomial(6,2)= 15

p(2)= 7.5/15

= 1/2

p(0)=m/n,

m=0

p(0)=0

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2 years ago

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