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3 years ago

LAST QUESTION thank you so much to nintendo who has been helping me with all of my questions!

Mathematics

1 answer:

melomori [17]3 years ago

3 0

Using the properties of arcs and inscribed angles, B is 110/2=55.

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Which of the following statements is NOT TRUE when finding the length of the missing leg BC?

aalyn [17]

You can’t let “c” represent the missing leg

5 0

2 years ago

Read 2 more answers

Solve for y: x/y - 8 = w

inysia [295]

X/y-8=w
x-8y=wy
x=wy+8y
x=y(w+8)
y= $\frac{x}{w+8}$

4 0

3 years ago

A 15-minute TV program consists of one commercial of 1

Margarita [4]

Answer:

6 7/8 minutes

Step-by-step explanation:

Say the entertainment segments are each x minutes long. Then, we have the equation:

x + x + 1 1/4 = 15

2x + 1 1/4 = 15

2x = 15 - 1 1/4

We need a common denominator. 1 1/4 = 5/4 and 15 = 60/4, so we now have:

60/4 - 5/4 = 55/4. Then: 2x = 55/4

Divide by 2 from both sides:

(55/4) / 2 = 55/8 = 6 7/8

Thus, each entertainment segment is 6 7/8 minutes long.

Hope this helps!

7 0

3 years ago

Read 2 more answers

What is the area, in square inches, of a circle with a diameter equal to 16 inches?

Lostsunrise [7]

Answer: 201.06 in²

Step-by-step explanation:

Use the equation πr² to find out the area of a circle.

Since the diameter is 16 the radius is half of that, 8.

π(8)²

Work out that equation and you're done!

7 0

3 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

3 0

3 years ago