Find values x and y

kondor19780726 [428]

Hi Annabanna11!
You can draw a model for 0.78 by drawing 100 parts and coloring 78 of them! 0.78=78/100=78%
I hope this helps:)

6 0

4 years ago

Read 2 more answers

How is the quotient 80,000÷2000 different from the quotient 80,000÷200 or 80,000÷20

Olegator [25]

2,000, 200 and 20 are similar except for the number of zeros.
You can remove a zero from each to equal the number of zeros in the divisor. So 80,000 ÷ 2,000 is equivalent to 80 ÷ 2 = 40 you just remove the 3 zeros
80,000 ÷ 200 is equivalent to 800 ÷ 2 = 400 you just keep removing 0s like for instance this time it was 2 lastly 80,000 ÷ 20 only allows us to remove 1 zero 8,000 ÷ 2 = 4,000. The smaller the divisor the greater the quotient when dividing the same number like for instance in this example 80,000

3 0

3 years ago

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and

charle [14.2K]

Answer:

A) 0.5737

B) 0.9884

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 160 lb and a standard deviation of 27.5 lb i.e.; $\mu$ = 160 lb and $\sigma$ = 27.5 lb

(A) We know that Z = $\frac{X - \mu}{\sigma}$ ~ N(0,1)

Let X = randomly selected pilot

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( $\frac{X - \mu}{\sigma}$ < $\frac{201 - 160}{27.5}$ ) = P(Z < 1.49) = 0.9319

P(X <= 150) = P( $\frac{X - \mu}{\sigma}$ < $\frac{150 - 160}{27.5}$ ) = P(Z < -0.3636) = P(Z > 0.3636) = 0.3582

Therefore, P(150 < X < 201) = 0.9319 - 0.3582 = 0.5737 .

(B) We know that for sampling mean distribution;

Z = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

P(X bar < 201) = P( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{201 - 160}{\frac{27.5}{\sqrt{39} } }$ ) = P(Z < 9.311) = 1 - P(Z >= 9.311)

= 0.999995

P(X bar <= 150) = P( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{150 - 160}{\frac{27.5}{\sqrt{39} } }$ ) = P(Z < -2.2709) = P(Z > 2.2709)

= 0.0116

Therefore, P(150 < X bar < 210) = 0.999995 - 0.0116 = 0.9884

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

8 0

3 years ago

Find the perimeter of the rectangle with the following vertices. (−6, −2), (0, −10), (5, 2), (−1, 10) 23 52 46 40

Licemer1 [7]

Answer:

46

Step-by-step explanation:

See attached for reference

The points given:

(−6, −2), (0, −10), (5, 2), (−1, 10)

They form a rectangle as seen in the picture.

We can notice that this is a parallelogram, as respective lines have same difference of coordinates.

So calculating only the two of the sides will be sufficient to get its perimeter:

a = √(-1+6)² + (10+2)² = √25+144 = √169= 13
b = √(0+6)² + (-10+2)² = √36+64 = √100 = 10

So, the perimeter:

P = 2(13+10) = 46

7 0

3 years ago

If you select one card at random from a standard deck of 52 cards, what is the probability that the card is a five AND a club?

horsena [70]

Probability is expressed as

number of favorable outcomes/number of total outcomes

The number of cards in a deck is 52(total outcomes)

Recall, there is only 1 five of clubs in a deck of cards(favorable outcomes)

Thus, the probability that the card is a five AND a club is 1/52

4 0

2 years ago