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Alex Ar [27]
2 years ago
5

Micah rows his boat on a river 4.48 miles downstream, with the current, in 0.32 hours. He rows back upstream the same distance,

against the current, in 0.56 hours. Assuming his rowing speed and the speed of the current are constant, what is the speed of the current?
Mathematics
2 answers:
ira [324]2 years ago
7 0
The speed of the current that Micah rows his boat on would be 3 miles per hour assuming that he is rowing 4.48 miles downstream with the current, in 0.32 hours.
Brilliant_brown [7]2 years ago
4 0

The answer is he is going 3 miles per hour

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4. What can you infer about tax law in a state such<br> as Indiana, Connecticut, or Michigan?
iragen [17]

Answer:

Y’all cpensive

Step-by-step explanation:o think

5 0
2 years ago
Tyler ate x fruit snacks, Han ate 3/4 less than that. Write an equation to represent the relationship between the number Tyler a
Citrus2011 [14]

Answer:

0.75x = y

or

1/4 = y

4 0
2 years ago
Aaron, Blaine, and Cruz are solving the equation 4 7 (7 − n) = −1. Aaron started his solution by multiplying both sides of the e
hjlf

Answer:

D. All three chose a valid first step toward solving the equation.

Step-by-step explanation:

Aaron, Blaine, and Cruz are solving the equation 4/7 (7 − n) = −1. Aaron started his solution by multiplying both sides of the equation by 7/4 . Blaine started by using the distributive property to multiply 4/7 by both 7 and −n. Cruz started by dividing both sides of the equation by 4/7 .

Which of the following is true?

A. Blaine and Cruz made an error in picking their first steps.

B. Cruz made an error in picking his first step.

C. All three made an error because the right side equals -1.

D. All three chose a valid first step toward solving the equation.

Given:

4/7(7 - n) = -1

Aaron:

4/7(7 - n) = -1

Multiple both sides by 7/4

4/7(7 - n) * 7/4 = -1 * 7/4

7 - n = -7/4

- n = -7/4 - 7

- n = (-7-28)/4

- n = -35/4

n = 35/4

Blaine:

4/7(7 - n) = -1

4/7(7 - n) × 7 = -1 × 7

4(7 - n) = -7

28 - 4n = -7

-4n = -7 - 28

- 4n = - 35

n = -35/-4

n = 35/4

Cruz:

4/7(7 - n) = -1

Divide both sides by 4/7

4/7(7 - n) ÷ 4/7 = -1 ÷ 4/7

4/7(7 - n) × 7/4 = -1 × 7/4

7 - n = -7/4

- n = (-7-28)/4

- n = -35/4

n = 35/4

D. All three chose a valid first step toward solving the equation.

7 0
2 years ago
One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples
JulijaS [17]

Answer:

Step-by-step explanation:

Hello!

To test if boys are better in math classes than girls two random samples were taken:

Sample 1

X₁: score of a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: Score in the calculus of a girl

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.

To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.

Then you can calculate the interval as:

[(X[bar]_1-X[bar_2) ± t_{n_1+n_2-2;1-\alpha /2} * Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11

t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708

[(82.3-81.2) ± 1.708* (6.11*\sqrt{\frac{1}{15}+\frac{1}{12}  }]

[-2.94; 5.14]

Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.

I hope this helps!

3 0
2 years ago
Sara wants to find the input value that produces the same output for the functions represented by the tables.
Korolek [52]

Answer:

The input value that produces the same output value in both charts is 2.

Step-by-step explanation:

You are given two functions f(x)=-0.5x+2 and g(x)=2x-3 with tables

\begin{array}{cc}x&f(x)\\-3&3.5\\-2&3\\-1&2.5\\0&2\\1&1.5\\2&1\\3&0.5\end{array}

and

\begin{array}{cc}x&g(x)\\-3&\\-2&\\-1&\\0&\\1&\\2&\\3&\end{array}

First, fill in the second table:

g(-3)=2\cdot (-3)-3=-6-3=-9\\ \\g(-2)=2\cdot (-2)-3=-7\\ \\g(-1)=2\cdot (-1)-3=-5\\ \\g(0)=2\cdot 0-3=-3\\ \\g(1)=2\cdot 1-3=-1\\ \\g(2)=2\cdot 2-3=1\\ \\g(3)=2\cdot 3-3=3

Hence, the second table is

\begin{array}{cc}x&g(x)\\-3&-9\\-2&-7\\-1&-5\\0&-3\\1&-1\\2&1\\3&3\end{array}

The input value that produces the same output value in both charts is 2.

6 0
2 years ago
Read 2 more answers
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