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3 years ago

Help me with 20 please

Mathematics

1 answer:

Murrr4er [49]3 years ago

8 0

24%------360(24% is 360m)

100%-------x (100% is x)

to find x we have too cross multiply

x=100*360/ 24= 1500

hope this helps

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This Stem-and-Leaf Plot represents the heights of the students on Ralph’s basketball team. One student’s height is missing from

kompoz [17]

Using the mean of the data-set, it is found that the missing height in the stem-and-leaf plot is given by:

D. 65 in.

The mean of a data-set is given by the <u>sum of all observations in the data-set divided by the number of observations</u>.

In this problem, following the key of the stem-and-left plot, there are 8 observations, given as follows:

54, 55, 56, 61, 63, 64, 70 and the missing value x.

The mean is of 61, hence:

$61 = \frac{54 + 55 + 56 + 61 + 63 + 64 + 70 + x}{8}$

423 + x = 61 x 8

x = 65.

Hence option D is correct.

More can be learned about the mean of a data-set at brainly.com/question/24628525

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2 years ago

Help me with dis pls

Iteru [2.4K]

Answer:

53, -84, 24, -38

Step-by-step explanation:

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3 years ago

A dormitory has n students, all of whom like to gossip. one of the students hears a rumor, and tells it to one of the other n −

Alekssandra [29.7K]

Answer:

$P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}$

Step-by-step explanation:

From the question, we have the following condition:

$p_1=p_2=1,\:and\:p_n=0$

We know that each student who hears the rumor tells it to a student picked at random from the dormitory (excluding, of course, himself/herself and the person from whom he/she heard the rumor)

The 3rd student can therefore tell the rumour to n-2 students but only n-3 will accept it.

$\implies p(3)=\frac{n-3}{n-2}$

Consequently, the 4th student must not tell the 1st and second students.

$\implies p(4)=\frac{n-3}{n-2}\times \frac{n-4}{n-2}$

We can rewrite this to observe a pattern:

$\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^2}$

$\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^{4-2}}$

$\implies p(r)=\frac{(n-3)(n-4)(n-5)...(n-r)}{(n-2)^{r-2}}$

Hence, the probability that the rumor is told r times without coming back to a student who has already is:

$P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}$

See attachment for complete question

4 0

4 years ago

a fish taco from Ama taco truck cost $4.50. a soda cost $1.25. if corey eats a fish taco and soda and wants to tip 15% how much

S_A_V [24]

Answer:

4.50 + 1.25 = 5.75 x.15 = 0.8625

Round that to .86

$5.75 +.86(tip)

.86 cents

Step-by-step explanation:

8 0

3 years ago

. The time required for a technician to machine a specific component is normally distributed with a mean of 2 hours and a standa

erma4kov [3.2K]

Answer:

a) There is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

b) There is a 0.42% probability that the technician will require at least 2.75 hours to complete one component

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.

In this problem, we have to be careful. The mean is in hours, while the standard deviation is in minutes. I am going to work with both in hours, as the problem states. 17 minutes is 0.283 hours, so:

$\mu = 2, \sigma = 0.283$

(a.) What is the probability that the technician can machine one component in 1.5 hours or less?

This probability is the pvalue of the Zscore when $X = 1.5$ . So:

$Z = \frac{1.5 - 2}{0.283}$

$Z = \frac{-0.5}{0.283}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

(b.) What is the probability that the technician will require at least 2.75 hours to complete one component?

The pvalue of the score of $X = 2.75$ is the probability that the technican will require less than 2.75 hours to complete one component. The probability that he will require at least 2.75 hours to complete one component is 1 subtracted by this pvalue. So:

$Z = \frac{2.75 - 2}{0.283}$

$Z = \frac{0.75}{0.283}$

$Z = 2.65$

$Z = 2.65$ has a pvalue of 0.99598.

This means that the probability that the technican will require at least 2.75 hours to complete one component is 1 - 0.99598 = 0.0042 = 0.42%.

4 0

4 years ago