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Rebecca has 21 math problems to solve. She has solved 2-7 of them. How many problems has she solved

Blizzard [7]

She has solved 6 problems. 21/7 times 2 equals 6.

7 0

3 years ago

Write an equation to represent the proportional relationship between the # of Vegetable Plants (x) and the # of Flowers (y).

yawa3891 [41]

Answer:

Proprortional Value is equal to 2/3 or 2:3

Step-by-step explanation:

4/6=2/3

6/9=2/3

8/12=2/3

10/15=2/3

14/21=2/3

6 0

3 years ago

You may need to use the appropriate technology to answer this question. A survey was conducted to determine whether hours of sle

Marina CMI [18]

Answer:

1) The null and alternative hypotheses.

H0: Hours of sleep per night is independent of age.

Ha: Hours of sleep per night is not independent of age.

Test Statistics χ²=4.009

The corresponding p-value for the test at χ²= 4.009 is 0.2604

Step-by-step explanation:

1) The null and alternative hypotheses.

H0: Hours of sleep per night is independent of age.

Ha: Hours of sleep per night is not independent of age.

2) The data is

≤ 39 ≥ 40 Total

6 hours 36 34 70

6.9-7 hours 60 57 117

7- 7.9 hours 79 77 156

8 hours or more 65 92 157

Total 240 260 500

3) The expected values are computed as follows

260×70/500=36.4

240×117/500 =56.16

260×117/500 =60.84

240×156/500 =74.88

260×156/500 =81.12

240×157/500=75.36

260×157/500 =81.64

4) The squared distances are calculated as

{(36 - 33.6)^2}/{ 33.6} = 0.171

{(34 - 36.4)^2}{ 36.4} = 0.158

{(60 - 56.16)^2}/{ 56.16} = 0.263

{(57 - 60.84)^2}/{ 60.84} = 0.242

{(79 - 74.88)^2}/{ 74.88} = 0.227

{(77 - 81.12)^2}/{ 81.12} = 0.209

{(65 - 75.36)^2}/{ 75.36} = 1.424

{(92 - 81.64)^2}{ 81.64} = 1.315

5) The significance level is set at α=0.05 for df = (4 - 1) (2 - 1) =3

degrees of freedom.

6) The critical region for this test is χ² = χ² >7.815

7) Test Statistics

χ²=i=∑(O ij −Eij )2/Eij

=0.171+0.263+0.227+1.424+0.158+0.242+0.209+1.315

= 4.009

8) Conclusion

As the calculated value χ² =4.009 does not lie in the critical region ≤χ² =7.815, null hypothesis is not rejected.

Therefore, we conclude that there is not enough evidence to accept the alternate hypothesis , at the α=0.05 significance level.

The p-value for the test at χ²= 4.009 is 0.2604

5 0

3 years ago

Factor the expression. 9y^2– 36

Nata [24]

Answer:

Answer: 9 (y - 2) (y + 2)

Step-by-step explanation:

Factor the following:

9 y^2 - 36

Factor 9 out of 9 y^2 - 36:

9 (y^2 - 4)

y^2 - 4 = y^2 - 2^2:

9 (y^2 - 2^2)

Factor the difference of two squares. y^2 - 2^2 = (y - 2) (y + 2):

Answer: 9 (y - 2) (y + 2)

5 0

3 years ago

The quality-control department of Starr Communications, the manufacturer of video-game cartridges, has determined from records t

notsponge [240]

Answer: (A) The probability that a cartridge purchased will have a video or audio defect is 1.9%

(B) The probability that a cartridge purchased will not have a video or audio defect is zero.

Step-by-step explanation: The data given shows that 1.2% (or 120) cartridges have video defects, 0.9% have audio defects (or 90) and 0.2% (or 20) have both audio and video defects.

The possible outcomes for all events (audio defects and video defects) is derived as 120 plus 90 which is equals 210 possibilities (or possible outcomes).

Therefore the probability of having an audio defect is calculated as follows;

P(Audio) = Number of required outcomes/Number of all possible outcomes

P(Audio) = 90/210

P(Audio) = 3/7

Also the probability of having a video defect is derived as follows;

P(Video) = Number of required outcomes/Number of all possible outcomes

P(Video) = 120/210

P(Video) = 4/7

However we should take note of the fact that 0.2% or 20 of the cartridges in the sample size has both audio and video defects. Hence the probability that a cartridge has both audio and video defects is calculated as;

P(Audio and Video) = Number of required outcomes/Number of all possible outcomes

P(Audio and Video) = 20/210

P(Audio and Video) = 2/21

To calculate the probability that a cartridge bought would have either an audio or a video defect would mean to add both probabilities together, but we MUST SUBTRACT the probability of having both an audio defect and video defect (that is P{Audio and Video}). The reason is that this is already included in both probabilities and we need to avoid double counting. Hence we have;

(A); P(Video OR Audio defect) = P(Audio) + P(Video) - P(Audio and Video)

P(Video OR Audio defect) = (3/7 + 4/7) - 2/21

P(Video OR Audio defect) = 1 - 2/21

P((Video OR Audio defect) = 19/21

Therefore the probability that a cartridge purchased will have a video or audio defect is 190, or better still 1.9%.

(B): From all possibilities shown, which is 210 possibilities of either events, we have determined that 120 will be the probability of having an audio defect and 90 will be the probability of having a video defect. Therefore the probability that a cartridge purchased will not fall into any of either possibilities is zero.

6 0

3 years ago