What time is 5 3/4 hours after 11:32 pm?

I need help with this math question I don’t understand

Illusion [34]

What’s your question bro

7 0

3 years ago

A café owner is designing a new menu and wants to include a decorative border around the outside of her food listings. Due to th

kipiarov [429]

The 13-in. by 9-in. rectangle where the food listings fit has an area of 13 in. * 9 in. = 117 in.^2

Adding 48 in.^2 for the border, the total area of the menu with the border will be 117 in.^2 + 48 in.^2 = 165 in.^2

The border has to have uniform width around the menu. We need to find the width of the border. Let the border be x inches wide. Then since you have a border at each of the 4 sides, the border will add 2x to the length of the rectangle and 2x to the width of the rectangle. The menu will have a length of 2x + 13 and a width of 2x + 9. The area of the larger rectangle must by 165 in.^2. The area of a rectangle is length times width, so we get our equation:

(2x + 13)(2x + 9) = 165

Multiply out the left side (use FOIL or any other method you know):

4x^2 + 18x + 26x + 117 = 165

4x^2 + 44x + 117 = 165

4x^2 + 44x - 48 = 0

Divide both sides by 4.

x^2 + 11x - 12 = 0

Factor the left side.

(x + 12)(x - 1) = 0

x + 12 = 0 or x - 1 = 0

x = -12 or x = 1

The solution x = -12 is not valid for our problem because the width of a border cannot be a negative number. Discard the negative solution.

The solution is x = 1.

Answer: The border is 1 inch wide.

Check. Add 2 inches to the length and width of the food listings rectangle to get 15 inches by 11 inches. A = 15 in. * 11 in.= 165 in.^2. Now subtract the area of the border, 48 in.^2, 165 in.^2 = 48 in.^2 = 117 in.^2, and you get the area of the 13-in. by 9-in. rectangle. This shows that our solution is correct.

6 0

3 years ago

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Tarush is a landscape architect. For his first public project, he is asked to create a small scale drawing of a garden to be pla

svet-max [94.6K]

Answer: 7.5 is the height, 5.0 is the base

5 0

3 years ago

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Pneumonoultramicroscopicsilicovolcanoconiosis how many letters

Jet001 [13]

Pneumonoultramicroscopicsilicovolcanoconiosis has 45 letters.

7 0

2 years ago

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Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste

Vladimir [108]

Answer:

The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

Let $\mu$ = true mean transmission of pages.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 10 pages {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, $H_A$ : $\mu$ > 10 pages {means that the true mean is greater than 10}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 14.44 pages

s = sample standard deviation = 4.45 pages

n = sample of fax transmissions = 35

So, test statistics = $\frac{14.44-10}{\frac{4.45}{\sqrt{35} } }$ ~ $t_3_4$

= 5.9028

(a) The value of t test statistics is 5.9028.

Now, at 0.01 significance level the z table gives critical values of 2.441 at 34 degree of freedom for right-tailed test.

Since our test statistics is more than the critical values of z as 5.9028 > 2.441, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_3_4$ > 5.9028) = Less than 0.05% {using t table)

7 0

3 years ago