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3 years ago

eight students share 5 blocks of modeling clay equally. What fraction of one block of modeling clay does each student get?

Mathematics

1 answer:

quester [9]3 years ago

8 0

5/8 trust me because when you divide 5 by 8 you get a number that is equal to 5/8

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Modeling the number of transport bus breakdowns in every month to determine if an additional bus should be purchased. The runnin

skelet666 [1.2K]

Answer:

The probability of exactly 2 breakdowns next month = 0.08422

Step-by-step explanation:

To calculate the probability of exactly 2 breakdowns next month, we use Poisson distribution formula.

P(X = x) = (e^-λ)(λˣ)/x!

Mean = λ = 5 breakdowns per month

x = 2 breakdowns per month

P(X = 2) = (e⁻⁵)(5²)/2!

P(X = 2) = 0.08422

6 0

3 years ago

If you have thirty -five (35) members in an organization that meets nine (9) times a year, how often will a member have to host

Olegator [25]

Answer:

the host will continue the way they are prepared.

3 0

3 years ago

I need help with this problem! Must explain why.

OLEGan [10]

Answer:

95°

Step-by-step explanation:

Angle 1 and 4 are opposite angles, thus equal, so:

3x+10 = 4x-15 =>

x = 25

Angle 1 and 4 are 85°

Angle 4 and angle 6 are called allied (or co-interior) angles, which are supplementary, i.e., angle 6 = 180 - angle 4, so angle 6 = 95°.

Angle 6 and 7 are again opposite, so angle 7 is 95° as well!

7 0

3 years ago

Use the diagram to find the measure of the given angle.

Lyrx [107]

Answer:

Step-by-step explanation:

4 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago