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Leviafan [203]
3 years ago
12

Misleading advertising messages regarding fast food are often given by __________. A. politicians B. doctors C. athletes D. dent

ists Please select the best answer from the choices provided. A B C D
SAT
1 answer:
Luda [366]3 years ago
3 0

Answer:

They are often given by athletes.

So the answer is C

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37 cos(x2) dx 0 Do the following. (a) Find the approximations T8 and M8 for the given integral. (Round your answer to six decima
dusya [7]

The approximations T8 and M8 for the given integral are:

  • T8 = 33.386321; and
  • M8 = 33.50794

<h3>What is an integral?</h3>

An Integral is a variable of which a given function is the derivative, i.e. it gives that function when differentiated and may express the area under the curve of the function's graph.

<h3>What is the explanation to above answer?</h3>

Given:

F(x) = 37 cox (x²)

Internal = [0,1] n = 8 in Δ x = 1/8

The sub intervals are:

[0, 1/8], [1/8, 2/8], [2/8, 3/8], [ 3/8, 4/8], [ 4/8, 5/8], [ 5/8, 6/8], [6/8, 7/8], [ 7/8, 1]

The mid points are given as:

1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16

and X₀ = 0, X₁ = 1/8, X₂ = 2/8

Using the Trapezium Rule which states that:

\int\limits^1_0 cos(x)^{2} } \, dx = Δx/2 [f(xo) + 2f(x1) 2f(x2) + ....+ 2f(x7) + f(x8)]

= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]

= 0.902333

Now

T8 = \int\limits^1_0 {37Cos(x)^{2} } \, dx

= 37\int\limits^1_0 {(0.902333)} } \, dx

= 37 (0.902333)

T8 = 33.386321

It is to be noted that the midpoints rule is given as;

\int\limits^1_0 {Cos(x)^{2} } \, dx  = Δx [f(1/16) + (3/16) + .... + f(15/16)]

= 1/8[f(1/16) + f (3/16) + f(5/16) + f(7/16) + f(9/16) + f(11/16) + f(13/16) + f(15/16)]

= 0.905620

From the above,

M8 = \int\limits^1_0 {37 Cos(x)^{2} } \, dx

= 37\int\limits^1_0 {Cos(x)^{2} } \, dx

= 37 (0.905620)

M8 = 33.50794

Learn more about integral at;
brainly.com/question/19053586
#SPJ1

6 0
2 years ago
An ice cream truck owner has determined that his maximum revenue occurs when he sells 1,250 cones per month. For every cone abov
devlian [24]

0.15|c-1,250|=250 Is it right?

8 0
3 years ago
Read 2 more answers
When dana found out her proposal was rejected, she stormed into her boss's office and yelled at him. Dana was experiencing emoti
Lisa [10]

Answer:

Emotional hijacking occurs when emotions take over the rational...

Explanation:

7 0
2 years ago
Y = x^2 - 5x + 3
Eduardwww [97]

Answer:

x = 6 (A)

Explanation:

We plug y = x^2 - 5x + 3 into 5x + y = 39 and we have

5x + x^2 - 5x + 3 = 39

-> x^2 = 36

-> x = 6

8 0
3 years ago
A survey found that​ women's heights are normally distributed with mean
zheka24 [161]

Answer:

a. 99.30% of the woman meet the height requirement

b.  If all women are eligible except the shortest​ 1% and the tallest​ 2%, then height should be between 58.32 and 68.83

Explanation:

<em>According to the survey</em>, women's heights are normally distributed with mean 63.9 and standard deviation 2.4

a)

A branch of the military requires​ women's heights to be between 58 in and 80 in. We need to find the probabilities that heights fall between 58 in and 80 in in this distribution. We need to find z-scores of the values 58 in and 80 in. Z-score shows how many standard deviations far are the values from the mean. Therefore they subtracted from the mean and divided by the standard deviation:

z-score of 58 in= z=\frac{58-63.9}{2.4} = -2.458

z-score of 80 in= z=\frac{80-63.9}{2.4} = 6.708

In normal distribution 99.3% of the values have higher z-score than -2.458

0% of the values have higher z-score than 6.708. Therefore 99.3% of the woman meet the height requirement.

b)

To find the height requirement so that all women are eligible except the shortest​ 1% and the tallest​ 2%, we need to find the boundary z-score of the

shortest​ 1% and the tallest​ 2%. Thus, upper bound for z-score has to be 2.054 and lower bound is -2.326

Corresponding heights (H) can be found using the formula

2.054=\frac{H-63.9}{2.4}  and

-2.326=\frac{H-63.9}{2.4}

Thus lower bound for height is 58.32 and

Upper bound for height is 68.83

8 0
3 years ago
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