Two-thirds of Tyrone's books are mystery novels. Tyrone has 42 books How many mystery novels does he have ?

Which graph represents a linear function? A coordinate plane is shown with a way line beginning along the x axis and continuing

Sliva [168]

-2-2 + 2 = 0(-2, 0)-1-1 + 2 = 1(-1, 1)00 + 2 = 2(0, 2)11 + 2 = 3(1, 3)22 + 2 = 4(2, 4)

5 0

2 years ago

Read 2 more answers

Plz someone help me ASAP

Morgarella [4.7K]

2 because if you divided 8/12 by 2/6=2

8 0

2 years ago

Lazar drives to work every day and passes two independently operated traffic lights. The probability that both lights are green

Natalija [7]

Conditional probability is a measure of probability of an event given that another event has occurred.
P ( A\ B ) = P ( A ∩ B ) / P ( B ) - the conditional probability of A given B
P ( A ∩ B ) = 0.41; P ( B ) = 0.59
P ( A \ B ) = 0.41 / 0.59 = 0.6949152 ≈ 0.69
Answer: B ) 0.69

7 0

2 years ago

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A charity fair raised $6,000 by selling 500 lottery tickets. There were two types of lottery tickets: A tickets cost $10 each, a

julsineya [31]

x= A ticket
y=b ticket
x+y=500
10x+60y=6000
-------------------
10x+60y=6000
- x+y=500 (*10) ---> 10x+10y=5000
-------------------
10x+60y=6000
- 10x+10y=5000
-------------------
50y=1000
y=1000/50
y=20
b ticket =20

x+y=500
x+20=500
x=500-20
x=480
a ticket=480
answer is the last option ,

6 0

2 years ago

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

2 years ago