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mafiozo [28]
3 years ago
15

What is the value of XY/z if x=1 y=-10 and z=2 A.5 B.-5 C.1/5 D.-1/5

Mathematics
2 answers:
Oliga [24]3 years ago
6 0

The answer is B. If you substitute the values of x, y and z in you get 1 * -10 which just is -10. -10 / 2 = -5.

Mashutka [201]3 years ago
5 0

Hi Xxmirzxxx,

Solution:

XY/z

=\frac{x * y}{z}

=\frac{1 * -10}{2}

=\frac{-10}{2}

= -5

Final Answer: -5

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marshall27 [118]
B is the right hope it help
8 0
3 years ago
Anyone help me wiv this question please
Jlenok [28]

Answer:

A: 8x-12

B: 5(n+3)

Step-by-step explanation:

A) Distribute 4 to the numbers inside the parenthesis.

4 x 2x= 8x

4 x -3= -12

8x-12

B) 5 is divisible to 15, so you can divide it with both numbers.

5/5=1

15/5=5

5(n+3)

3 0
3 years ago
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If you roll a fair six-sided die and a fair four-sided die, what is the probability that the dice add to 6 or higher
Lena [83]
Okay so you have to figure out all the possible ways to get six or higher. between a 6 sided dice and a 4 sided dice. We do 6*24 to find all the possibilities.

The probability of getting a 6 or higher is 14/24



4 0
3 years ago
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A graduate class consists of six students. What is the probability that at least 5 of them are born either in April or in Octobe
nlexa [21]

Answer:

0.0671% probability that at least 5 of them are born either in April or in October

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they were born in April or October, or they were not. The probabilty of a student being born in April or October is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of a student being born in April or October

April has 30 days, October 31

The year has 365 days. So

p = \frac{61}{365} = 0.167

A graduate class consists of six students.

This means that n = 6

What is the probability that at least 5 of them are born either in April or in October?

P(X \geq 5) = P(X = 5) + P(X = 6)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{6,5}.(0.167)^{5}.(0.833)^{1} = 0.000649

P(X = 6) = C_{6,6}.(0.167)^{6}.(0.833)^{0} = 0.000022

P(X \geq 5) = P(X = 5) + P(X = 6) = 0.000649 + 0.000022 = 0.000671

0.0671% probability that at least 5 of them are born either in April or in October

8 0
3 years ago
What is the answer 2×2​
Arlecino [84]
The answer to that is 4.
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3 years ago
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