I need help. Thanks so much

Order the numbers from greatest to least.<br><br>4. 234,358; 23,208; 23,098

coldgirl [10]

Answer:

23,098; 23,208 ; 234,358

Step-by-step explanation:

Hope this helps

6 0

3 years ago

PLS HELP RN <br> $4000 student loan. 10% compound interest annually. Total loan amount by 4 years?

ddd [48]

Answer:

$2,400

Step-by-step explanation:

We get 10% of 4,000 by dividing 4,000 by 10.

4,000 ÷ 10 = 400

To get the amount that 4,000 is subtracted by we multiply 400 by 4.

400 × 4 = 1,600

To get the answer we subtract 4,000 by 1,600.

4,000 - 1,600 = 2,400

Hope I helped!

3 0

3 years ago

If f(x) = 5x - 11, find f(-2m + 5)

sukhopar [10]

$f(-2m + 5)=5(-2m+5)-11=-10m+25-11=-10m+14$

8 0

3 years ago

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:

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