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iren [92.7K]
3 years ago
11

Find the least common multiple. 15 and 3 = CO.CC

Mathematics
2 answers:
tankabanditka [31]3 years ago
5 0
The least common multiple for 15 and 3 is 15
Savatey [412]3 years ago
3 0

Answer:

15: <u>15</u>, 35, 50, 65 <u> </u><u> </u><u> </u><u>LCM= 15</u><u> </u><u> </u>

3: 3, 6, 9, 12, <u>15</u>

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When g equals- 3 answer is 1

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The coordinates of the vertices of triangle XYZ are given in the table.
SCORPION-xisa [38]

X': (0,-6)
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Read 2 more answers
The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day
Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(between 236 and 281 days)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%

b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

1-\dfrac{1}{(2)^2} = 75\%

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0
3 years ago
PLS HELP (LAST QUESTION)
rosijanka [135]

Answer:

You would have $18

Step-by-step explanation:

Have a good day!

4 0
3 years ago
A train travels 95 kilometers in 1 hours, and then 97 kilometers in 1 hours. What is its average speed?
aleksandrvk [35]
Average speed is given by [Total Distance] ÷ [Total Time]

Average speed = \frac{95+97}{1+1} = \frac{192}{2}=96 km/hour
8 0
3 years ago
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