All categories

2 years ago

Help 35 points Which value of c markes

Mathematics

2 answers:

miskamm [114]2 years ago

7 0

X=6 because 6-3 is 3 then which makes 3/4 . 3/4 also equals 9/12 plus 2/3 (which equals 8/12) is 17/12

sveticcg [70]2 years ago

3 0

Answer:

x=6 Isolate the variable by dividing each side by factors that don't contain the variable.

You might be interested in

PLEASEEEEE HELP I DO REALLY NEED IT

Mkey [24]

Answer:

the second option

Step-by-step explanation:

the SSS congruence postulate would be used

4 0

2 years ago

Combine the like terms to create an equivalent expression −n+(−3)+3n+5

spin [16.1K]

Answer:

2n+2

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

irakobra [83]

First one is 14
Second one is 45
Third one is 43
Fourth one is 40

8 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

If a 5-pound bag of potatoes costs $5.25 , then how much would 7 pounds cost?

sashaice [31]

Answer:

7.35

Step-by-step explanation:

5 pounds -> 5.25

1 pound -> 1.05

7 pounds-> 7.35

4 0

2 years ago