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ivolga24 [154]
3 years ago
6

Please help me on 21 and 22

Mathematics
1 answer:
Trava [24]3 years ago
4 0
A1= 2520
a2= 14
a3= 9 
a4= 20
b1= 6426
b2= 27
b3= 7 
b4= h
b5= 34
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Estimate the difference. Use benchmarks with decimal parts of 0, 0.25, 0.50, or 0.75.
Firdavs [7]
Your real exact answer would be 0.29 but a bench mark would be 0.25 so B
3 0
3 years ago
I WILL IVE U 5 STARS
matrenka [14]

Answer:

  A. 1 rectangle, 2 triangles

  B. AB = AE = 5

  C. 36.5 square units

Step-by-step explanation:

<h3>A.</h3>

The attached figure shows 1 rectangle (square) and two triangles.

__

<h3>B.</h3>

These sides are aligned with the grid, so their length is simply the difference in coordinates along the line:

  AB = 2 -(-3) = 5

  AE = 3 -(-2) = 5

__

<h3>C.</h3>

The area of the square is ...

  A = s^2 = 5^2 = 25

The area of triangle BCF is ...

  A = 1/2bh = 1/2(3)(5) = 15/2

The area of triangle CDE is ...

  A = 1/2bh = 1/2(8)(1) = 4

The total area is the sum of the areas of the square and two triangles:

  total area = 25 +7.5 +4 = 36.5 . . . square units

_____

<em>Additional comment</em>

We note that segment CE divides the figure into <em>trapezoid</em> ABCE and <em>triangle</em> CDE. The trapezoid has bases 5 and 8, and height 5, so its area is ...

  A = 1/2(b1 +b2)h = 1/2(5 +8)(5) = 32.5

Triangle CDE has the same area as computed above, 4 square units. So, the total area of the figure is ...

  32.5 +4 = 36. 5 . . . . square units

4 0
3 years ago
How do you find the domain of a function?
nadezda [96]
To find the domain of a function you have to find all the possible input values.
4 0
3 years ago
At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
dmitriy555 [2]

Answer:

The interval is [910.053; 959.946]

p-value 0.00596

Decision: Reject null hypothesis.

Step-by-step explanation:

Hello!

You need to make a 95% Confidence Interval for the population mean of scholarship examination scores for the freshman.

It is known to be μ= 900 and the assistant dean wants to test if it changed.

The study variable is:

X: -scholarship examination score of one applicant.

population variance is known as σ²= (180)²

Assuming that the variable has a normal distribution the formula for the interval is:

X[bar] ± Z_{1-\alpha /2}*\frac{S}{\sqrt{n} }

935 ± 1.96*\frac{180}{\sqrt{200} }

The interval is [910.053; 959.946]

To test if the examination scores have changed the hypothesis is:

H₀: μ = 900

H₁: μ ≠ 900

α: 0.05

To use a Confidence Interval the following conditions should be met:

1) Both the test and the interval should be made for the same parameter.

2) The hypothesis has to be two_tailed

3) Confidence level 1 - α and significance level α should be complementary.

To make the decision you have to see if the value given to the population mean in the null hypothesis is contained or not by the interval.

If the value is contained by the interval, you do not reject the null hypothesis.

If the value is not contained by the interval, then the decision is to reject the null hypothesis.

Since 900 is not contained by the 95% Confidence interval [910.05; 959.95], the decision is to reject the null hypothesis. This means that the scholarship examination scores of freshman applications have changed.

To calculate the p-value you have to know the value of the statstic under the null hypothesis:

Z= \frac{935-900}{\frac{180}{\sqrt{200} } }= 2.749 ≅ 2.75

p-value is

P(Z<2.75) + P(Z>2.75)= P(Z<2.75) + (1 - P(Z<2.75))= 0.00298+ (1 - 0.9702= 0.00596

I hope it helps!

4 0
3 years ago
HELP ASAP FOR BRAINLIEST ANSWER!!!
velikii [3]

Answer:

14

Step-by-step explanation:

You'd subtract the angles. 68 - 54 is 14. Hope this helped. <3

6 0
3 years ago
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