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Luda [366]
3 years ago
6

The height of a trapezoid is 8 in. and its area is 80 in2. One base of the trapezoid is 6 inches longer than the other base. Wha

t are the lengths of the bases? Complete the explanation of how you found your answer.

Mathematics
1 answer:
joja [24]3 years ago
4 0

Answer:40

Step-by-step explanation:

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Need help on 1 and 7 please giving out 19 points
Serjik [45]

Answer:

for number one the answer is no for 7 the answer is 5 seconds

Step-by-step explanation:

4 0
4 years ago
Marco is making a kite, using sticks for its diagonals. One stick is 2 feet long and the other is 3 feet long. How much material
Marat540 [252]

Answer:

Marco will need 3\ ft^{2} of material to make the kite

Step-by-step explanation:

we know that

To know how much material Marco will need to make the kite, the area must be calculated.

Remember that the area of the kite is equal to

A=\frac{1}{2}[d1*d2]

where

d1 and d2 are the diagonals of the kite

we have

d1=2\ ft

d2=3\ ft

substitute

A=\frac{1}{2}[2*3]=3\ ft^{2}

3 0
4 years ago
Read 2 more answers
Whats992,449 rounded to the nearest hundred thousand
Andreas93 [3]
Yep, the answer is 1,000,000. To explain the process of how the answer is figured out, the 9 next to the 9 in the hundred thousand's place tells us that we need to round up. So, we'd be rounding up to 1,000,000. Hope this helped you understand it :)
3 0
3 years ago
Read 2 more answers
Find the range and interquartile range for the data represented by the box plot.
Anarel [89]

<u>Given</u>:

Given that the data are represented by the box plot.

We need to determine the range and interquartile range.

<u>Range:</u>

The range of the data is the difference between the highest and the lowest value in the given set of data.

From the box plot, the highest value is 30 and the lowest value is 15.

Thus, the range of the data is given by

Range = Highest value - Lowest value

Range = 30 - 15 = 15

Thus, the range of the data is 15.

<u>Interquartile range:</u>

The interquartile range is the difference between the ends of the box in the box plot.

Thus, the interquartile range is given by

Interquartile range = 27 - 18 = 9

Thus, the interquartile range is 9.

3 0
3 years ago
Read 2 more answers
A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We wan
svlad2 [7]

Answer:

t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322

Step-by-step explanation:

We have a sample of size n = 35, \bar{x} = 92 and s = 5. As we want to construct a 95% confidence interval to estimate the value of the true population mean \mu, we should use the pivotal quantity given by T = \frac{\bar{X}-\mu}{S/\sqrt{n}} which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322, i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that P(-2.0322\leq T\leq 2.0322) = 0.95 because of the simmetry of the t distribution.

7 0
3 years ago
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