Calculate the Midpoint M of line segment AB given, A(4,-10) and B(8,4).

give you 300 dollars if u get this right. you have my word The lifetimes of 20,000 light bulbs are normally distributed. The mea

stira [4]

Wow ! I didn't know the numbers when I read the question,
but with that much riding on it, I had to go look them up and
learn something. You probably could have done the same thing
and saved yourself some money, but it's too late for that. It's all
mine now, as soon as I finish writing this answer.

In the normal distribution, 68.27% of the values lie within one
standard deviation of the mean, 95.45% lie within two standard
deviations, and 99.73% lie with three standard deviations.

Within one sigma: 68.27% of 20,000 = 13,654
13,654 light bulbs will last between 190 days and 270 days.

Within two sigma: 95.34% of 20,000 = 19,068
19,068 light bulbs will last between 150 days and 310 days.

Here's the one you asked for:

Within three sigma: 99.73% of 20,000 = 19,946
19,946 light bulbs will last between 110 days and 350 days.

Only 54 light bulbs will burn out in less than 110 days
OR keep going more than 350 days.

OK. Now you owe me.

6 0

4 years ago

What is the least number that should be subtracted from 1385 to get a perfect square Also find the square root of the perfect sq

Artemon [7]

Answer:

16 and the square root of 1369 is 37

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious

Dominik [7]

Answer:

Part1) $z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191$

$p_v =2*P(Z>2.191)=0.0284$

Readjustment is needed

Part 2) $z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$

$z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$

$p_v =2*P(Z$

No readjustment is needed

Step-by-step explanation:

Data given and notation

Part 1

$\bar X=16.32$ represent the sample mean

$\sigma=0.8$ represent the population standard deviation

$n=30$ sample size

$\mu_o =16$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu =16$

Alternative hypothesis : $\mu \neq 16$

Compute the test statistic

The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191$

Give the appropriate conclusion for the test

Since is a two tailed test the p value would be:

$p_v =2*P(Z>2.191)=0.0284$

Using the value of $\alpha =0.05$ we see that $pv$ and we have enough evidence to reject the null hypothesis.

Readjustment is needed

The rejection zone is given by:

$(-\infty ;-1.96) , (1.96;\infty)$

Part 2

$\bar X=15.82$ represent the sample mean

We can replace in formula (1) the info given like this:

$z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$

Give the appropriate conclusion for the test

Since is a two tailed test the p value would be:

$p_v =2*P(Z$

Using the value of $\alpha =0.05$ we see that $pv>\alpha$ and we have enough evidence to FAIL to reject the null hypothesis.

No readjustment is needed

The rejection zone is given by:

$(-\infty ;-1.96) , (1.96;\infty)$

What action would you recommend ?

Review the procedure.

Do you reach the same conclusion ?

No we got different conclusions

5 0

4 years ago

8 friends want to share 6 pizzas equally. How much pizza will each friend get?

N76 [4]

Answer:

0.75

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

PLZ HELP!!! I Will give brainliest. What is the value of x in sin(3x)=cos(6x) if x is in the interval of 0≤x≤π/2

sertanlavr [38]

Answer:

sin(2x)=cos(π2−2x)

So:

cos(π2−2x)=cos(3x)

Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that

(π2−2x)=±3x

i)

π2=5x

x=π10

ii)

π2=−x

x=−π2

Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)

So we see that

(π2−2x−2π)=±3x

iii)

π2−2π=5x

x=−310π

iv)

π2−2π=−x

x=2π−π2=32π

Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)

So we get ∀n∈N

i) x=π10+2πn

ii) x=−π2+2πn

iii) x=−310π+2πn

(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)

So we conclude that there are really three solutions and then the periodic extensions of those three solutions.

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