All categories

3 years ago

Which fractions are closer to 0 than 1? Check all that apply.

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

6 0

Answer:

B. and D the rest are half or closer to 1

hope this helps

have a good day :)

Step-by-step explanation:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

b and d are closer to 0 than 1 mark me brainliest

Step-by-step explanation:

You might be interested in

What is Manuel percent grade if he got 23 correct out of 25?

olganol [36]

If he got a 23 out of 25 just divide 23 by 25 and your answer would be

Answer-92%

5 0

3 years ago

Read 2 more answers

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863$

$P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734$

$P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113$

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

5 0

3 years ago

Combine like terms. 6x + 4 – 3x + 5

Ivahew [28]

X = 3

Combine like terms: -4 + -5 = -9
-9 + 6x + -3x = 0
Combine like terms: 6x + -3x = 3x
-9 + 3x =0

Solving -9 + 3x = 0

Solving for variable ‘x’

Moving all terms containing my x to the left, all other terms to the right
Add ‘9’ to each side of the equation -9 + 9 + 3x = 0 + 9

Combine like terms: -9 + 9 = 0
0+3x=0+9
3x = 0 + 9

Combine like terms: 0 + 9 = 9
3x = 9

Divide each side by ‘3’.
X = 3

SORRY THIS IS SO LONG

7 0

3 years ago

Three pounds of lawn seed covers 1800 square feet. How many 4 pound bags are needed to cover 8400 square feet?

g100num [7]

1800 ft² = 3 pounds

[divide by 3 through]
1800 ÷ 3 = 3 ÷ 3
600ft² = 1 pound

[8400 ÷ 600 = 14; so multiply by 14 through]
600 x 14 = 1 x 14
8400ft² = 14 pounds

we need 14 pounds to cover 8400 square feet.

[Find number of 4 pounds bags needed]
14 ÷ 4 = 3.5
So we need to use 4 bags of 14 pounds of lawn seeds.

7 0

3 years ago

Find the value of y and simplify completely

Law Incorporation [45]

Here we have three similar triangles

In similar triangles sides are in proportion

The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse.

$\frac{part of hypotenuse }{altitude} =\frac{altitude}{other part of hypotenuse}$