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2 years ago

11

A jet plane traveling at 550 mph overtakes a propeller plane traveling at 200 mph that had a 4 hour head start. How far from the

starting point are the planes?

1 answer:

Orlov [11]2 years ago

8 0

The plane traveling 200 mph, started 4 hours earlier.

As at when they meet, let the 550 mph, have traveled for x hours.

The 200 mph would have traveled (x + 4) hours, since it started 4 hours earlier.

As at when they met, the distance covered is the same:

Distance = Speed * hour

550x = 200(x + 4)

550x = 200x + 200*4

550x = 200x + 800

550x - 200x = 800

350x = 800

x = 800/350

x = 16/7 hours.

The distance travelled = 550x = 550 * 16/7 = 8800/7 miles

The distance = 8800 / 7 miles or ≈ 1257.14 miles

Hope this explains it.

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Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

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3 years ago

Dose Congruent Mean Having The Same Measure

Julli [10]

So Congruent. Two figures are congruent if they have the same shape and size. Two angles are congruent if they have the same measure. Two figures are similar if they have the same shape but not necessarily the same size.

I hope that help u:)

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3 years ago

Two of the lights at the local stadium are flickering. They both just flickered at the same time. One of the lights flickers eve

irinina [24]

1,665
because 777+888 = 1,665

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3 years ago

Read 2 more answers

The area of a rectangle is 229.2 m2. If the length is 15 m, what is the perimeter of the rectangle

igor_vitrenko [27]

Answer:

$\large\boxed{P=60.56\ m}$

Step-by-step explanation:

The formula of an area of a rectangle:

$A=lw$

l - length

w - width

We have

$A=229.2\ m^2\\\l=15\ m$

Substitute:

$15w=229.2$ <em>divide both sides by 15</em>

$w=15.28\ m$

The formula of a perimeter of a rectangle:

$P=2(l+w)$

Substitute:

$P=2(15+15.28)=2(30.28)=60.56\ m$

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3 years ago

For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr

pshichka [43]

Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables $X_i$ with the following distribution:

$X_i Bin (1,p) = Be(p)$ bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

$Z = \sum_{i=1}^N X_i$

From the info given we know that $N \sim Bin (M,q)$

We need to proof that $Z \sim Bin (M, pq)$ by the definition of binomial random variable then we need to show that:

$E(Z) = Mpq$

$Var (Z) = Mpq(1-pq)$

The deduction is based on the definition of independent random variables, we can do this:

$E(Z) = E(N) E(X) = Mq (p)= Mpq$

And for the variance of Z we can do this:

$Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2$

$Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2$

And if we take common factor $Mpq$ we got:

$Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]$

And as we can see then we can conclude that $Z \sim Bin (M, pq)$

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3 years ago