All categories

2 years ago

I'm hosting a charity event fo r42 people. One out of every 7 guests would love a cupcake.

Mathematics

2 answers:

GuDViN [60]2 years ago

8 0

Answer:

There's two ways to solve this problem. #2 is for after you solved it the first way.

$42 / 7 = 6$
$6*7 = 42$

the 2nd way is the first problem in reverse.

Greeley [361]2 years ago

6 0

Answer:

6. (If I understand the qeution)

Step-by-step explanation:

42 divided by 7 = 6

You might be interested in

Can someone please help me with this question!!!

Elena-2011 [213]

Answer:

Step-by-step explanation:

x+102=4x+6

102=3x+6

96=3x

32=x

Hope this helps!!!

5 0

3 years ago

Find - 2 2/6 + (5/6)

Bad White [126]

Answer:-3/2

Step-by-step explanation:

-2 2/6+5/6

-7/3 + 5/6

(2x-7+1x5) ➗ 6

(-14+5) ➗ 6

-9 ➗ 6

-9/6=-3/2

5 0

3 years ago

The length of a rectangle is 4 units shorter than half the width. If the length of the rectangle is 18 units, which equation can

never [62]

Answer:

The width is 11 units.

Step-by-step explanation:

First, you have to add 4 to 18 which is 22. Then divide by 2.

8 0

3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago

Can you graph a line if its slope is undefined? Explain.

JulijaS [17]

You can but it would be going straight up and down

6 0

4 years ago