At 3 pm jamie leaves her house

Use the net to compute the surface area of the three-dimensional figure.

Vlada [557]

The answer is 246 units²

Surface area= 2(lw + lh + wh)

3 0

3 years ago

Read 2 more answers

What is the solution for m >7/12

Over [174]

Answer:

Inequality Form: m > 7/12

Interval Notation: (7/12, ∞)

4 0

3 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = the number that are not correctly calibrated.

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = 0.35

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = 0.54

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = 0.11

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago

John asked 60 people to name their favourite fruit here are his results

LuckyWell [14K]

Pears 15-30
Bananas 20-40

5 0

3 years ago

Please hellllppppppp!!!!

Anna11 [10]

The difference is going to be 7

4 0

3 years ago