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Arisa [49]
3 years ago
5

A train is traveling 55 mph to a destination that is 330 mi away. The function f(x)=−55x+330f(x)=−55x+330 represents the remaini

ng number of miles to the destination after x hours of travel. What is the practical domain of the function f?
Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
F(x) = - 55x + 330

The domain is the set of possible input-values, i.e. x-values.

You know that f(x) cannot bebigger than 330 not lower than zero, then construct the inequality:

0 ≤ f(x) ≤ 330

0 ≤ - 55x + 330 ≤ 330

Multilply by - 1 and change the inequality signs:

0 ≥ 55x - 330 ≥ -330

Add 330 to all the members

330 ≥ 55x ≥ 0

Divide by 55

6 ≥ x ≥ 0 => 0 ≤ x ≤ 6

The the possible x-values (practical domain) is all the real numbers between 0 and 6 inclusive.
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The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years
inn [45]

Answer:

b. The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

c. Student's t distribution, as it is a two-sample test for the difference between means.

d. Test statistic t = 5.42.

P-value = 0

e. They appear to be significantly different, as the sample data gives strong evidence that the difference is greater than 0.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean life spans in the county is significantly different for whites and nonwhites.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

The significance level is 0.05.

The sample 1 (white), of size n1=124 has a mean of 45.3 and a standard deviation of 12.7.

The sample 2 (non-white), of size n2=82 has a mean of 34.1 and a standard deviation of 15.6.

The difference between sample means is Md=11.2.

M_d=M_1-M_2=45.3-34.1=11.2

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12.7^2}{124}+\dfrac{15.6^2}{82}}\\\\\\s_{M_d}=\sqrt{1.301+2.968}=\sqrt{4.269}=2.066

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{11.2-0}{2.066}=\dfrac{11.2}{2.066}=5.42

The degrees of freedom for this test are:

df=n_1+n_2-2=124+82-2=204

This test is a two-tailed test, with 204 degrees of freedom and t=5.42, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>5.42)=0.0000002

As the P-value (0.0000002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean life spans in the county is significantly different for whites and nonwhites.

4 0
3 years ago
Help fast hshshdhwbebsolamehvrve
rosijanka [135]

Answer:

Slope: -3/2

Y intercept: 4

Slope intercept form: y=-3/2x+4

Step-by-step explanation:

:) I hope this is good for you

5 0
2 years ago
35 more than twice the value of e is 65
almond37 [142]


here is what I think since 2 multiplied by 15 equals 30 you at 35 more to that 30 and you get the answer 60. Is that how it is supposed to go?


5 0
3 years ago
U divided by 25 =13 please do step by step and show work thanks :))
Anna71 [15]
Answer U/25=13
             U=13 divided 1/25
             U=13 multiply 25
             U=325
6 0
3 years ago
In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2
Ne4ueva [31]

Answer:

The probability that on any given day the water supply is inadequate 0.1991

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

P(X>9)= 1-P(X\leq 9)\\

Expanding the function and putting the given values, we get -

[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991

The probability that on any given day the water supply is inadequate 0.1991

5 0
3 years ago
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