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FrozenT [24]
3 years ago
6

Solve for z -p(d + z) = -2+59​

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Answer: z = (57/ -p) - d

Step-by-step explanation:

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How do I find X? This is a quadrilateral inside of a circle.
soldi70 [24.7K]

Answer:

circle it?

Step-by-step explanation:

°

5 0
3 years ago
A survey of 250 households showed 26 owned at least one snow blower. Find a point estimate for p, thepopulation proportion of ho
lions [1.4K]

Answer:

The population proportion of households that own at least one snow blower is 0.104.

Step-by-step explanation:

In order to find the population proportion of households that own at least one snow blower, you have to divide the number of people that said they owned at least one snow blower by the number of households surveyed:

26/250=0.104

According to this, the answer is that a point estimate for p, the population proportion of households that own at least one snow blower is 0.104.

3 0
3 years ago
A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on four numbers pays 8 to 1 (that is, if you bet $1 and one of t
Natasha_Volkova [10]

Answer:

Lose $0.05

Step-by-step explanation:

There are 38 possible spots on the roulette wheel (numbers 1 to 36, 0 and 00).

If the player can choose four numbers on single $1 bet, his chances of winning (W) and losing (L) are as follows:

P(W) = \frac{4}{38} \\P(L) = 1-P(W) = 1-\frac{4}{38} \\P(L) = \frac{34}{38}

The expected value of the bet is given by the probability of winning multiplied by the payout ($8), minus the probability of losing multiplied by the bet cost ($1)

EV=\frac{4}{38}*\$8 -\frac{34}{38}*\$1\\EV= -\$0.05

On each bet, the player is expected to lose 5 cents ($0.05).

4 0
3 years ago
What is 1 over 4 subtract 4 over 5
fomenos
....go to the school for this lol kids
3 0
3 years ago
There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit
Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a \frac{3}{11} chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is \frac{8}{11}, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is\frac{2}{11}, and the probability of picking the second red ball is \frac{1}{10} and the probability of picking the green ball is\frac{1}{9}. We want to multiply thm again, as per the multiplication rule like the last problem.

\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}

5 0
3 years ago
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