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3 years ago

Counting from 1 to 46, how many 4s will you encounter? 8 10 11 12 14

Mathematics

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

Step-by-step explanation:

4=1

14=2

24=3

34=4

40=5

41=6

42=7

43=8

44=9

45=10

46=11

You will encounter 11 4s.

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12. Carter has 3 statues of various sizes in his garden. Statue A is one foot taller than

enot [183]

Answer: B is 11 feet tall

A=x+1 since A is one more than B

C=x+1/2 since C is one half more than B

B=x since B is just B

A+B+C=34.5

x+1+x+x+1/2=34.5

3x+1.5=34.5

3x=33

x=11

B is 11 feet tall

7 0

2 years ago

Abby wants to enclose a rectangular area of 20 square feet to use as a garden. Write a function that could be used to find the g

Thepotemich [5.8K]

Xy=20 As the width gets smaller, the length should get larger.

6 0

3 years ago

PLEASE HELP!!

I am Lyosha [343]

36 + 18 = 54

(1) 18/54 = 1/3 = 33%

43 + 24 = 67

(2) 24/67 = 36%

I hope these are right

3 0

3 years ago

Read 2 more answers

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

valina [46]

Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

$P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

$P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1} + \frac{9}{20} * \frac{9- 1}{20- 1}$

$P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19} + \frac{9}{20} * \frac{8}{19}$

$P(Same)=\frac{20}{380} + \frac{30}{380} + \frac{72}{380}$

$P(Same)=\frac{20+30+72}{380}$

$P(Same)=\frac{122}{380}$

$P(Same)=\frac{61}{190}$

4 0

3 years ago

Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six

baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

$P(A|d) =\frac{P(A \cap d)}{P(d)}$

$P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}$

$P(A|d)= 0.3$

Hence, the probability that a part was manufactured on machine A is 0.3

8 0

3 years ago