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4 years ago

Segment ab and segment bc are tangent to circle d. find x if ab = 5x + 11 and bc = 3x + 25.

Mathematics

1 answer:

nadya68 [22]4 years ago

3 0

A 5x/b +11/b and b=3x/c+25/c

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Select each expression that shows a correct method for finding 36% of 400.

kvv77 [185]

The answer is 136 136=36% of 400

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4 years ago

-10 F is colder than -4 degrees?

Aleks04 [339]

Answer: yes, it is a smaller temperature, which means that is "colder"

Step-by-step explanation:

I guess that the question is:

Is -10°F colder than - 4°C?

Ok, when we have a temperature T in Celcius, the equation to transform this quantity to the Fahrenheit scale is:

T' = (T*9/5) + 32°

Replacing T by the temperature in celcius, we get:

T' = (-4°*9/5) + 32° = 24.8°F

This means that -4°C is equivalent to 24.8°F

And -10°F is a smaller value than 24.8°F (which mean that is colder)

This implies that -10°F is colder than -4°C

Then the statement is true.

4 0

3 years ago

A problem states: "There are 7 more girls than boys in a club. There are 23 members in the club in all. How many girls are there

dlinn [17]

Answer:

Step-by-step explanation:

Number of boys = g - 7.

g - 7 + g = 23

2g = 30

g = 15.

The are 15 girls in the class.

3 0

3 years ago

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Len [333]

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7 0

3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

3 years ago