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2 years ago

Select the reason why these triangles are similar. If they are not, select “not similar”

Mathematics

1 answer:

larisa [96]2 years ago

8 0

Answer:

B. SAS

Step-by-step explanation:

3/1= 2.25/075

The 2 sides are congruent and the angle between them is common

B. SAS is the answer

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Use the property of exponents to rewrite the expression (-4qr)(-4qr)(-4qr)(-4qr)

ivanzaharov [21]

Answer:

(-4qr)^4

Step-by-step explanation:

(-4qr)(-4qr)(-4qr)(-4qr)

There are 4 sets of -4qr being multiplied together

(-4qr)^4

5 0

2 years ago

for two functions, a(x) and b(x), a statement is made that a(x) = b(x) at x=2. what is definitely true about x=2

Musya8 [376]

Answer:

Functions a(x) and b(x) intersect each other at x = 2.

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3 years ago

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The expression 3(x-9) is equivalent to (Look at picture for answer choices)

Vaselesa [24]

Answer:

The last one

Step-by-step explanation:

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2 years ago

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13. A bakery is having a buy 4, get 1 free sale on cakes. This sale is eguivalent to what percentage off the regular price of al

exis [7]

Answer: 20

Step-by-step explanation:

3 0

2 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

2 years ago