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3 years ago

Consider the figures shown below. Find the values of ∠1 and ∠2 for the given figures and complete the table accordingly.

Mathematics

1 answer:

iragen [17]3 years ago

5 0

Answer:

Figure 1: 98, 82

Figure 2: 64

Figure 3: 157, 23

Step-by-step explanation:

Figure 1: complementary angles theorem, coordinating angles theorem

Figure 2: vertical angles theorem

Figure 3: alternate interior angles theorem, complementary angles theorem

Let me know if this helped!

**please mark brainliest!!!

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A boy makes a cuboid of plasticine of sides 5cm, 2cm, 5cm. How manny such cuboids will he need to form a cube?

ruslelena [56]

Answer:

Step-by-step explanation:

L.C.M. of 5,2=10

so it should be of dimensions 10×10×10

it should have 5 cuboids in first row.

then it will make a cuboid 5×10×5

5 cuboids in the second row on top of first row .

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double the cuboids to make the length=10 and height=10..

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3 years ago

You have a party at your house. mAfter the party, you have 0.50.5 liters of juice remaining. How many milliliters of juice do yo

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Answer:

A liter is 1000 mL so half of 1000 = 500

500 mL is the answer!Step-by-step explanation:

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2 years ago

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Jerry is going to paint a circle in his living room. If the radius is 8 inches, what is the area that Jerry needs to cover with

adelina 88 [10]

Answer:201.06

Step-by-step explanation:

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3 years ago

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Akimi4 [234]

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3 years ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

2 years ago