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3 years ago

Which expression is equivalent to 15 + 35

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:

It says “which” so is it multiple choice??

Also a picture would help!!

You might be interested in

Which expression is equivalent to 5^3√6c+7^3√6c, if c ≠ 0

scZoUnD [109]

Answer:

Option C

Step-by-step explanation:

complete question

A. 35[3]6cB. 12[3]12cC. 12[3]6cD.72c

The given equation can be written as

5^3√6c+7^3√6c

5 * $\sqrt{3 *3*2c}$ + 7 * $\sqrt{3 *3*2c}$

5 * 3 * $\sqrt{2c}$ + 7 * 3 * $\sqrt{2c}$

15 * $\sqrt{2c}$ + 21 * $\sqrt{2c}$

36 * $\sqrt{2c}$

12 [3]* $\sqrt{2c}$

Option C is correct

4 0

3 years ago

I dont know what this is no links please

Gnoma [55]

Answer:

Hamm i think si r or q mm sorry if it's wrong

7 0

3 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago

Suppose that a box of 86 circuits is sent to a manufacturing plant. Of the 86 circuits shipped, 5 are defective. The plant manag

Triss [41]

Answer:

the probability that the shipment is accepted is 0.8865

Step-by-step explanation:

Given the data in the question;

N = 86, n/d = 5 and n = 2

now, without replacement

the probability that the shipment is accepted will be;

probability that the shipment is accepted = probability that non is defectives

so p( non is defective ) = ( (86-5)/86) × ((86-5-1)/(86-1))

p( non is defective ) = ( 81 / 86) × (80/85)

p( non is defective ) = 0.8865

Therefore, the probability that the shipment is accepted is 0.8865

7 0

3 years ago

Please help ASAPPPPP

iren [92.7K]

Answer:The third box

Step-by-step explanation:

You dont have to write the steps

6 0

3 years ago