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Oksi-84 [34.3K]
4 years ago
9

Look at this rectangle

Mathematics
1 answer:
Stolb23 [73]4 years ago
8 0

the new area will be three times the old area

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-7=7d-8 solve for the variable d
kvv77 [185]

-7 = 7d - 8

Add 8 to both sides.

1 = 7d

Divide both sides by 7.

<h3>d = 1/7</h3>
3 0
3 years ago
Subtract. <br> Express your answer in simplest form<br><br> 3 1/4 - 2 3/4
lord [1]
The answer to this is .5 or 1/2.
3 0
4 years ago
Read 2 more answers
Given f(x) = 4(1 – x), what is the value of f(-10)
Lubov Fominskaja [6]

Answer:

44

Step-by-step explanation:

The function formula is given as

f(x) = 4(1 – x)

To find the f(-10) we will substitute into the function formula, then we have

f(-10)= 4[1-(-10)]

=4[11]

=44

Hence, the value of f(-10)= 44

8 0
3 years ago
Square root of 22801 by long division method
mezya [45]
First step: partition the number you want to square root into a block of 2 digits, starting from the last digit (first diagram)

Second step: As our number is a five-digits, we ends up with 2  28  01. Pick a number that could be squared to get the first partition, 2. This number is 1, since 1×1=1

Third step: Write 1 on the top and on the side, as shown in the second graph

Fourth step: Double the number on the side, which is 1+1=2 and use this number as the first digit for the next multiplier. Meanwhile, subtract 1 from 2 inside the root sign to get 1, then pull the other two digits, 28

Fifth step: We need a value in the boxes that when we multiply together will give a number less than 128. We choose 5 as 25×5=125

Sixth step: Subtract 125 from  128 to give 3, and as the same concept with long division, bring down the 0 and the 1. So we have 301

Seventh step: Add 5 to the multiplier on the left, so 20+5=30, which we will use on the side as the hundred and ten digits.

Final step: Find a number to fit in the boxes. We choose 1 since 301×1=301

And hence the square root of 228801 is 151

3 0
3 years ago
Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t
Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let \sigma_{1}^{2} be the variance for the population of weight gains for rats given a low dose, and \sigma_{2}^{2} the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2} vs H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}. We have that the sample standard deviation for n_{2} = 22 female control rats was s_{2} = 28 g and for n_{1} = 18 female low-dose rats was s_{1} = 51 g. So, we have observed the value

F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176 which comes from a F distribution with n_{1} - 1 = 18 - 1 = 17 degrees of freedom (numerator) and n_{2} - 1 = 22 - 1 = 21 degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0
3 years ago
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