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2 years ago

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Four friends share the profit of a lawn care service equally. The friends charge $45 per lawn and spend $30 of their total earni

ngs on gasoline for the lawnmowers each week. Which inequality can be used to find m, the number of lawns they need to mow next week so that each person earns at least $275? A. 45m - 30/4 _> 275 B. 45m/4 - 30 _>275 C. 45m - 30/4 _< 275 D. 45m/4 - 30 _< 275

2 answers:

Pavel [41]2 years ago

7 0

Answer:

The answer is actually A) 45m-30/4 ≥ 275

Step-by-step explanation:

It says, "of the total earnings", so after they took off the 30 dollars, they split it between themselves. Hope this helps ;)

Mkey [24]2 years ago

5 0

Answer:

I'm pretty sure the answer is B

Step-by-step explanation:

If it's wrong, I'm super sorry. But have a nice day❤. Peace out✌

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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.

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HELLO PLEASE GIVE ME THE RIGHT ANSWER ASAP THANK YOU !

seropon [69]

$\large\underline{\sf{Solution-}}$

$\textsf{Given expression;}\\$

$\rm{23 = \frac{7}{9}(6x - 36) + 9 } \\$

$\rm{\longmapsto} \: 23 = \frac{(42x - 256)}{9} + 9 \\$

$\rm{\longmapsto} \: 23 = \frac{(42x - 256 + 81)}{9} \\$

$\rm{\longmapsto} \: 23 = \frac{(42x - 175)}{9} \\$

$\rm{\longmapsto} \: \frac{23}{1} = \frac{(42x - 175)}{9} \\$

$\textsf{By doing cross multiplication, we get}\\$

$\rm{\longmapsto}23(9) = 1(42x - 175) \\$

$\rm{\longmapsto}207 =42x - 175\\$

$\rm{\longmapsto} \: 42x = - 175 - 207 \\$

$\rm{\longmapsto} \: 42x = - 379 \\$

$\rm \therefore \: x = - \frac{375}{42} \\$

$\textsf{Hence, the value of x will be -375/42 respectively.}\\$

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