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3 years ago

Stephen does not like yogurt very much, but he loves

1 answer:

8 0

5+2+3=10 total snacks
5/10 or 1/2 are yogurts so 1/2 is the chance everyday
1/2×1/2×1/2=1/8 this is the chance Stephen gets yogurt 3 consecutive days so 1-1/8=
7/8 chance he doesn't get yogurt three consecutive days

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HELLO PLEASE HELP!! HAVE A GOOD DAY/NIGHT!!

aliina [53]

So, XZ= 7x+1 and PZ= 4x-1. To figure out XP we have to subtract those two.

(7x+1)-(4x-1)

3x-2

3(8)-2

24-2

XP=22

6 0

3 years ago

Read 2 more answers

Line segment AB has vertices A(-3, 4) and B(1, -2). A dilation centered at the orgin is applied to AB. The image has vertices A(

bixtya [17]

Answer:

option 2

Step-by-step explanation:

consider the coordinates A (- 3, 4 ) and A' (- 1, $\frac{4}{3}$ )

since the dilatation is centred at the origin, then corresponding coordinates are multiples/ divisors of each other, then image to original gives scale factor.

scale factor = $\frac{A'}{A}$ = $\frac{-1}{-3}$ = $\frac{1}{3}$ and $\frac{\frac{4}{3} }{4}$ = $\frac{1}{3}$

similarly B (1, - 2 ) and B' ( $\frac{1}{3}$ , - $\frac{2}{3}$ )

$\frac{\frac{1}{3} }{1}$ = $\frac{1}{3}$ and $\frac{-\frac{2}{3} }{-2}$ = $\frac{1}{3}$

4 0

2 years ago

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.

yawa3891 [41]

Answer:

The probability is 0.003

Step-by-step explanation:

We know that the average $\mu$ is:

$\mu=500$

The standard deviation $\sigma$ is:

$\sigma=100$

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

We seek to find

$P(x800)$

For P(x>800) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{800-500}{100}$

$Z=3$

The score of Z = 3 means that 800 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

$P(x>800)=0.15\%$

For P(x<200) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{200-500}{100}$

$Z=-3$

The score of Z = -3 means that 200 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

$P(x$