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2 years ago

The Hartman family is saving $400 monthly for Ronald's college education. The family anticipates they will need to

Mathematics

1 answer:

butalik [34]2 years ago

3 0

Answer:

D. The family will likely have enough money. They will have saved $19,200 and have accumulated interest.

Step-by-step explanation:

Got 100% on edge.

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Mr. mole left his burrow that lies 7 meters below the ground and started digging his way deeper into the ground, descending at a

Keith_Richards [23]

Answer:

A(t)=−1.8t−4.5

Step-by-step explanation:

Mr. Mole is descending at a constant rate, so we are dealing with a linear relationship.

We know that Mr. Mole descended at a rate of 1.81.81, point, 8 meters per minute, so the slope \green mmstart color green, m, end color green is \green{-1.8}−1.8start color green, minus, 1, point, 8, end color green, and our function looks like A(t)=\green{-1.8}t+\pink bA(t)=−1.8t+bA, left parenthesis, t, right parenthesis, equals, start color green, minus, 1, point, 8, end color green, t, plus, start color pink, b, end color pink.

We also know that after 555 minutes, Mr. Mole was 13.513.513, point, 5 meters below the ground, which means that A(5)=-13.5A(5)=−13.5A, left parenthesis, 5, right parenthesis, equals, minus, 13, point, 5. We can substitute this into the formula of the function to find \pink bbstart color pink, b, end color pink:

\qquad\begin{aligned}A(5)&=-13.5\\ \green{-1.8}\cdot5+\pink b&=-13.5\\ -9+\pink b&=-13.5\\ \pink b&=\pink{-4.5}\end{aligned}

A(5)

−1.8⋅5+b

−9+b

=−13.5

=−4.5

This means that Mr. Mole's initial altitude is 4.54.54, point, 5 meters below the ground.

AKA the answer is A(t)=−1.8t−4.5

3 0

2 years ago

A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.

From t table, at $\alpha= 0.01$ , reject null hypothesis if $t$ .

Since $t=1.986>-2.998$ , fail to reject null hypothesis.

Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.

5 0

1 year ago

Which function below has the largest slope?

postnew [5]

The slope of f(x) can be computed as
(f(4) -f(2))/(4 -2) = (12 -4)/(2) = 8/2 = 4

The slope of g(x) is the coefficient of x
10

The slope of h(x) is
(1 level)/(1 play) = 1

The function with the largest slope is g(x).

5 0

2 years ago

Read 2 more answers

The library charged Monica $50 for a book that was 10 days overdue

barxatty [35]

Erm what’s your question?

8 0

2 years ago

In a right triangle, angle a measures 20. The side opposite angle a is 10 centimeters long. Approximately how long is the hypote

AleksAgata [21]

sin(angle)=opposite leg/hypotenuse

sin(20) = 10/ hypotenuse