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3 years ago

8

Complete the remainder of the table

1 answer:

marissa [1.9K]3 years ago

5 0

Answer:

_3

_6/3+4=2

0=2*0/3+4=0

3=2*3/3+4=6

6=2*6/3+4

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HELP!!!!!!!!!!! What are two possible measures of the angle below?

Andreyy89

Answer: First Option -90° and 630°

Solution:

1) In clockwise the angle is negative, then is -90°

2) Counterclockwise the angle is positive and it could be:

270°

270°+360°=630°

Then, the answer is -90° and 630°

6 0

4 years ago

Gina earns $19,750 annually and has 20% of her gross earnings withheld. What is the net amount of her paycheck each pay period i

Kay [80]

Confuseing sorry can't answer that

8 0

3 years ago

Read 2 more answers

A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five

exis [7]

Answer:

$\dfrac{14}{29}$

Step-by-step explanation:

Let <em>P(A) </em>be the probability that goggle of type A is manufactured

<em>P(B) </em>be the probability that goggle of type B is manufactured

<em>P(E)</em> be the probability that a goggle is returned within 10 days of its purchase.

According to the question,

<em>P(A)</em> = 30%

<em>P(B)</em> = 70%

<em>P(E/A)</em> is the probability that a goggle is returned within 10 days of its purchase given that it was of type A.

P(E/B) is the probability that a goggle is returned within 10 days of its purchase given that it was of type B.

$P(A \cap E)$ will be the probability that a goggle is of type A and is returned within 10 days of its purchase.

$P(B \cap E)$ will be the probability that a goggle is of type B and is returned within 10 days of its purchase.

$P(E \cap A) = P(A) \times P(E/A)$

$P(E \cap A) = \dfrac{30}{100} \times \dfrac{5}{100}\\\Rightarrow P(E \cap A) = 1.5 \%$

$P(E \cap B) = P(B) \times P(E/B)$

$P(E \cap B) = \dfrac{70}{100} \times \dfrac{2}{100}\\\Rightarrow P(E \cap B) = 1.4 \%$

$P(E) = 1.5 \% + 1.4 \% \\P(E) = 2.9\%$

If a goggle is returned within 10 days of its purchase, probability that it was of type B:

$P(B/E) = \dfrac{P(E \cap B)}{P(E)}$

$\Rightarrow \dfrac{1.4 \%}{2.9\%}\\\Rightarrow \dfrac{14}{29}$

So, the required probability is $\dfrac{14}{29}.$

7 0

4 years ago

Find the length of a rectangle with a perimeter of 148 inches and width of 36 inches.<br> inches

Alexeev081 [22]

Answer:

Perimeter of rectangle = 2L + 2W

2L + 2 * 36 = 148

2L = 148 - 72

2L = 76

L = 38 inches

6 0

3 years ago

Read 2 more answers

How many seconds will light leaving Los Angeles take to reach the following locations (a) San Francisco (about 500km), (b) Londo

kondor19780726 [428]

Answer:

a) It takes 0.0017s for the light to reach San Francisco.

b) It takes 0.033s for the light to reach London.

c) It takes 1.334s for the light to reach Mars.

d) It takes 149.7s for the light to reach Venus.

Step-by-step explanation:

Here we can solve this problem by using this following formula:

$s = \frac{d}{t}$

In which s is the speed(in km/s), d is the distance(in km) and t is the time(in s).

The light speed is 299 792 458 m / s = 299,792.458 km/s, so $s = 299,792.458$

(a) San Francisco (about 500km)

Find t when $d = 500$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{500}{t}$

$299,792.458t = 500$

$t = \frac{500}{299,792.458}$

$t = 0.0017s$

It takes 0.0017s for the light to reach San Francisco.

b) London(about 10,000km)

Find t when $d = 10,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{10,000}{t}$

$299,792.458t = 10,000$

$t = \frac{10,000}{299,792.458}$

$t = 0.033s$

It takes 0.033s for the light to reach London.

(c) the Moon (400,000km)

Find t when $d = 400,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{400,000}{t}$

$299,792.458t = 400,000$

$t = \frac{400,000}{299,792.458}$

$t = 1.334s$

It takes 1.334s for the light to reach Mars.

(d) Venus (0.3 A.U. from Earth at its closest approach).

Each A.U. has 149,59,7871 km.

So 0.3 A.U. = 0.3*(149,597,871) = 44,879,361.3. This means that $d = 44,879,361.3$ . So:

$s = \frac{d}{t}$

$299,792.458 = \frac{44,879,361.3}{t}$

$299,792.458t = 44,879,361.3$

$t = \frac{44,879,361.3}{299,792.458}$

$t = 149.7s$

It takes 149.7s for the light to reach Venus.

3 0

3 years ago