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2 years ago

There are 14 pieces of chalk and 42 packages of

Mathematics

2 answers:

dem82 [27]2 years ago

7 0

Answer:

qjwjjwmwjwu 4f fsnfh the t4hhrnrhnn4

Phantasy [73]2 years ago

5 0

C.
14

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If a+3b=13 and a+b=5 the value of b is

hoa [83]

Answer:b=4

Step-by-step explanation:

7 0

3 years ago

Please anyone can give me it's urgent....

Pavlova-9 [17]

Step-by-step explanation:

From the figure it is clear that:

x+110°=180

x=180-110

x=70

For y:

the left exterior angle of line q is also70.

By the definition of alternate exterior angle:

y=70

Note:if you need to ask any question please let me know.

3 0

3 years ago

Does anyone know 1/4 x ? = -2?

SSSSS [86.1K]

Answer:

-8

Step-by-step explanation:

1/4 x = -2

You want the value of x. x is being multiplied by 1/4. To get x alone on the left side, you need to divide both sides by 1/4. Dividing by 1/4 is the same as multiplying by 4.

Multiply both sides of the equation by 4.

4 * 1/4 x = -2 * 4

4 * 1/4 = 1, so on the left side you get 1x which is just x.

x = 8

Answer: -8

8 0

3 years ago

Read 2 more answers

Why is it important to use the order of operations to evaluate algebraic experssions?

snow_tiger [21]

If you don’t your answer will not be correct also when you use them it helps you find the correct answer

7 0

2 years ago

Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

Step-by-step explanation:

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.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

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

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

8 0

4 years ago