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3 years ago

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A man places a mirror on the ground and sees the reflection of the top of a flagpole as shown in the figure below. The two trian

gles in the figure are similar. Find the height h of the flagpole. Note that the two triangles are proportional to one another. a = 3, b = 7, c = 28

1 answer:

tigry1 [53]3 years ago

5 0

Answer:

Step-by-step explanation:

B

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Determine whether the statement describes a descriptive or inferential statistic. A recent poll of 2935 luxury car owners in Wes

N76 [4]

Answer:

2300

Step-by-step explanation:

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3 years ago

55 36/72 rounded to the nearest whole number

Paraphin [41]

The answer would be 55.927... but rounded it would be 56.

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3 years ago

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le

frutty [35]

Answer:

a) Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}>120$

b) Calculate the statistic

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

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Step-by-step explanation:

Data given and notation

$\bar X_{1}= 8*60 +5 =485$ represent the mean for the new sample

$\bar X_{2}=5*60+40=340$ represent the mean for the old sample

$s_{1}=55$ represent the sample standard deviation for the new sample

$s_{2}=30$ represent the sample standard deviation for the old sample

$n_{1}=100$ sample size selected for the new sample

$n_{2}=120$ sample size selected for the old sample

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

t would represent the statistic (variable of interest)

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

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the new product has a battery life more than two hours longer than the leading product., the system of hypothesis would be:

Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}>120$

If we analyze the size for the samples both are higher than 30 but we don't know the population deviations so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})- \Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

b) Calculate the statistic

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

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

7 0

3 years ago

What is the range for the following numbers?<br> 5,1,12,3,4,2,10

Lubov Fominskaja [6]

Answer: 11......................................................................................

8 0

3 years ago

Read 2 more answers

As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur

Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math. It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

$y=Ce^{kt}$

Filling in our formula with the 2 conditions we are given:

$65=Ce^{10k}$ and $85=Ce^{15k}$

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k. If

$65=Ce^{10k}$ then

$\frac{65}{e^{10k}}=C$ which, by exponential rules is the same as

$C=65e^{-10k}$

Since that value of C is the same as the value of C in the other equation, we sub it in:

$85=65e^{-10k}(e^{15k})$

Divide both sides by 65 and use the rules of exponents again to get

$\frac{85}{65}=e^{-10k+15k}$ which simplifies down to

$\frac{85}{65}=e^{5k}$

Take the natural log of both sides to get

$ln(\frac{85}{65})=5k$

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

$65=Ce^{10(.0536527973)}$

which simplifies down to

$65=Ce^{.536527973}$

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

$y=38.01038064e^{(.0536527973)(23)}$ which simplifies a bit to

$y=38.01038064e^{1.234014338}$

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

6 0

3 years ago