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2 years ago

Can you please help me please I need help with this promblem

Mathematics

1 answer:

insens350 [35]2 years ago

8 0

Car B travels 10 more miles per hour than Car A, The line B intersects at 50, While line A intersects at 40 when they hit 1 hour! I Like this Math! Bring me back to this!!!

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Which choice is the appropriate simplified form n restriction for the variable 12x^2 - 8x / 4x^2

tatiyna

$\frac{12x^{2} - 8x}{4x^{2}} = \frac{4x(3x-2)}{4x^{2}} = \frac{3x-2}{x} ,where 

x \neq 0
\\ \\B. \frac{3x-2}{x} , x \neq 0$

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3 years ago

What is 450 increased by 75%

Pavel [41]

526.5 is the correct answer haha

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3 years ago

The scatter plot below represents the weight of a box of cookies based on the number of cookies in the box. A trend line shows t

Nana76 [90]

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3 years ago

Fei Yens dog eats 8 ounces of dog food each day. Fei Yen bought a 28 pound bag of dog food. how many 8 ounce servings are in a 2

steposvetlana [31]

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2 years ago

There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am

Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

$n = 479, \pi = \frac{130}{479} = 0.2714$

Hence the bounds of the interval are found as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112$

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0

2 years ago