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3 years ago

I need to know this please help me

Mathematics

1 answer:

lara [203]3 years ago

5 0

Answer:

x is 16

Step-by-step explanation:

Multiply the 2/5 by 8/8 and you should get 16/40

Hope this helped!

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Bill and Joe have 60 dollars between them. Bill has half as much as Joe. How much does Bill have?

Katyanochek1 [597]

Bill and Joe have 60 dollars between them. Bill has half as much as Joe.

so Bill = B and Joe = J

20 (B) + 40 (J) = 60

Bill has half the amount of Joe, and so 40/2 = 20

subtract 60 from 20 (the answer for Bill) and you get 40, (the answer for Joe)

hope this helps

6 0

3 years ago

What’s the answer does anyone know?

Alexus [3.1K]

56.57

Explanation:
In the process,
Diameter= 18,
Radius(r)=
18
2

∴
Radius= 9
Now,
Circumference (Perimeter) = ?
According to the formula,
Perimeter= 2
×
22
7
×
r

Taking the equation,
Perimeter= 2
×
22
7
×
r

2
×
22
7
×
9

396
7

56.57142857

56.57

Let's hope this helps you:)

8 0

3 years ago

Read 2 more answers

Help fast please I going to fail

Scilla [17]

Answer:

0.5
0.25
0.4
1.05
0.05
Hope this helps

4 0

2 years ago

vertices of parallelogram ABCD are A(2, 0), B(4, 3), C(3, 1). How many possibilities are there for the position of the fourth ve

nordsb [41]

Answer:

1 possibility is there for the the position of the fourth vertex.

7 0

3 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago