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galben [10]
4 years ago
5

What is 37 divided by 32.93

Mathematics
2 answers:
kenny6666 [7]4 years ago
4 0
The answer would be 1.1235
Masteriza [31]4 years ago
3 0
1.12359551

Good luck :)
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In this right triangle, you are given the measurements for the hypotenuse, c, and one leg, b. The hypotenuse is always opposite the right angle and it is always the longest side of the triangle.
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Number x Number y which, when multiplied by itself, equals number x Square root y
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3 years ago
Independent random samples from two regions in the same area gave the following chemical measurements (ppm). Assume the populati
Basile [38]

Answer:

d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.

Step-by-step explanation:

Hello!

You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.

Be X₁: Chemical measurement in region 1 (ppm)

Sample 1

n= 12

981 726 686 496 657 627 815 504 950 605 570 520

μ₁= 678

σ₁= 164

Sample mean X[bar]₁= 678.08

X₂: Chemical measurement in region 2 (ppm)

Sample 2

n₂= 16

1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844

μ₂= 812

σ₂= 239

Sample mean X[bar]₂= 811.94

Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.

Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.

[(X[bar]₁-X[bar]₂)±Z_{1-\alpha /2}* \sqrt{\frac{Sigma^2_1}{n1}+\frac{Sigma^2_2}{n_2}  }]

Z_{1-\alpha /2}= Z_{0.95}= 1.648

[(678.08-811.94)±1.648*\sqrt{\frac{164^2}{12}+\frac{239^2}{16}  }]

[-259.49;-8.23]ppm

Both bonds of the interval are negative, this means that with a 90% confidence level the difference between the population means of the chemical measurements of region 1 and region 2 may be included in the calculated interval.

You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.

I hope it helps!

6 0
4 years ago
On a certain​ route, an airline carries 8000 passengers per​ month, each paying ​$70.A market survey indicates that for each​ $1
son4ous [18]

Answer:

The maximum revenue is $661,250.

Step-by-step explanation:

The total revenue (R) is the product between the number of passengers (N) and the price of each ticket (P), so the inicial revenue is:

R = N * P = 8000 * 70 = $560,000

For each 1$ increase in the ticket, the airline loses 50 passengers, so if we call "x" the increase in the ticket price, we have that the revenue equation will be:

R = (8000-50*x) * (70+x) = 560000 + 8000*x - 50*70*x -50x2

R = -50*x2 + 4500*x + 560000

The value of x that gives the maximum value of a quadratic function is found using the formula:

x = -b/2a

where a and b are coefficients of the quadratic equation (in our case, a = -50 and b = 4500)

So the value of x that gives the maximum revenue is:

x = -4500 / (-100) = 45

Using this value in the revenue equation, we have that the maximum R is:

R = -50*45^2 + 4500*45 + 560000 = $661,250

(To get this revenue, the ticket will cost 70+45 = $115 and there will be 8000-50*45 = 5750 passengers)

3 0
3 years ago
Help please! it wold make my day :)
strojnjashka [21]

Answer: Kahn is wrong because the sequence increases by 5 after every number. The next two numbers in the sequence would be 28, 33. Kahn puts 29, 33 which is incorrect.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Which of the following points would be a solution to this system of linear inequalities? ​
pychu [463]

Let's plug in both x and y values in the equation and check if the inequality is true...

\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}

(4,1) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 4 and y as 1

1 \leqslant 4 + 1 \\ 1 <  -  \frac{4}{2}  - 1

0  \leqslant  4 + 1 \\ 1 <  - 2 - 1

0 \leqslant 5 \\ 1 <  - 3

  • <em>This is not a solution because both are not true, Only one equation is giving true as answer which is not correct!~</em>

\purple{ \rule{300pt}{3pt}}

(0, - 3) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 0 and y as -3

- 3 \leqslant 0 + 1 \\  - 3 <  -  \frac{0}{2}  - 1

- 3 \leqslant 1 \\  - 3 <  - 0 - 1

- 3 \leqslant 1 \\  - 3 <  - 1

  • <em>This is a solution because both are true, </em><em>Both the </em><em>equations</em><em> </em><em>are</em><em> </em><em>true</em><em> </em><em>which</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>ordered</em><em> </em><em>pair</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>solution</em><em>!</em><em>~</em>

<u>Thus</u><u>,</u><u> </u><u>Option</u><u> </u><u>B</u><u> </u><u>(</u><u>0</u><u>,</u><u>-</u><u>3</u><u>)</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>correct</u><u> </u><u>choice</u><u>.</u><u>.</u><u>.</u><u>.</u>

6 0
3 years ago
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