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olya-2409 [2.1K]
3 years ago
13

What’s the answer to this ??

Mathematics
1 answer:
lesya [120]3 years ago
6 0

Answer:

Step-by-step explanation:

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Amy class went the fair they saw 22 horses and 18 ponies they also saw 18 rabbits and 37 hares compare the the number of horses
Dima020 [189]
The answer is 15. 37 plus 18 equals to 55 and 22 plus 18 equals to 40. 55 minus 40 equals to 15.
8 0
4 years ago
P (-2,6), Q (9,6), R (7, -1), S (-4, -1) are the vertices of a quadrilateral.
vova2212 [387]

Step-by-step explanation:

Parallelogram

Reflect y=-1 ;x changes sign

6 0
3 years ago
Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a red card
joja [24]

Answer: \dfrac{3}{51}

Step-by-step explanation:

Given : The total number of cards in a deck = 52

Number of red cards = 26

There are two types of red  cards : diamond and heart.

Number of diamond cards = 13

The probability that the first card is a diamond :-

\dfrac{13}{52}=\dfrac{1}{4}

Since diamond is also a red card.

Now, the total cards left = 51

The number of red cards left = 12

The probability that the second card is a red card (without repetition) is given by :-

\dfrac{12}{51}

Now, the probability of choosing a red card for the second card drawn, if the first card, drawn without replacement, was a diamond :-

\dfrac{1}{4}\times\dfrac{12}{51}=\dfrac{3}{51}

7 0
4 years ago
Help please due today
Vadim26 [7]

Answer:

Line a. goes to table 3, line b. goes to table 2, and line c. goes to table 1.

Step-by-step explanation:

Here are the 3 lines graphed (I even labeled each for you) so you can have a bit of a visual.

Hopefully you can find the points on each graph.

(Hint: The x row represents the x coordinate of an ordered pair, and the y row represents the y coordinate of an ordered pair.)

Ordered pairs look like this btw (x,y)

Hope this helps :)

6 0
3 years ago
The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.
Wittaler [7]

Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so \mu = 75, \sigma = 8.6.

a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation

This is the pvalue of Z when X = 60. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 75}{8.6}

Z = -1.74

Z = -1.74 has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.

b) What’s the 99th percentile of the distribution of the number of hits?

What is the value of X when Z has a pvalue of 0.99.

Z = 2.35 has a pvalue of 0.99

So

Z = \frac{X - \mu}{\sigma}

2.35 = \frac{X - 75}{8.6}

X - 75 = 20.21

X = 95.21

The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) What’s the probability of observing between 80 and 90 hits an hour?

This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.

For X = 90

Z = \frac{X - \mu}{\sigma}

Z = \frac{90 - 75}{8.6}

Z = 1.74

Z = 1.74 has a pvalue of 0.95907

For X = 80

Z = \frac{X - \mu}{\sigma}

Z = \frac{80 - 75}{8.6}

Z = 0.58

Z = 0.58 has a pvalue of 0.71904

So

There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour

6 0
3 years ago
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