Find the value of x. Round to the nearest tenth. 40 X 7 x = [?]

3. The table gives the coordinates of a

marin [14]

Answer:

B

Step-by-step explanation:

Look at the X and Y coordinates carefully. X and Y coordinates are swapped and Y coordinate got negative sign.

4 0

2 years ago

HELLLLPP 6(X+7) for x=9. Answer choices/ 1. 12. 2. 42. 3. 96. 4. 16

Lina20 [59]

It's 96

6(9+7)

6(16)

6*16

=

96

7 0

2 years ago

Multiply (2.5 x 10 ^-8) ( 9x10 ^ - 10 ) Express answer in scientific notation

earnstyle [38]

2.25 x 10^-17. Because it's negative you move the decimal to the left. So 2.5 x 10^-8 is .000000025 and 9 x 10^-10 is .0000000009.

4 0

3 years ago

Read 2 more answers

What is the length of side a? Round to the nearest tenth of an inch

slamgirl [31]

Answer:

ok this is easy all it is 16-7=9 so a=9 so all it is 7x9=72 then its 72 divided by 2 so your answer is 36

3 0

2 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

2 years ago