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3 years ago

4(m+3)=36 what’s the value of m

Mathematics

2 answers:

alisha [4.7K]3 years ago

4 0

Answer:

the answer is 6

Step-by-step explanation:

lys-0071 [83]3 years ago

4 0

Answer:

m=6

Step-by-step explanation:

4(m+3) = 36

4m + 12 = 36

4m + (12 -12) = (36 -12)

4m = 24

4m/4 = 24/4

m = 6

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abruzzese [7]

We have for this case:
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3 years ago

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A recipe for fruit punch calls for 8 cups of Sprite for every 12 cups of fruit juice. Which equation can be used to find x, the

Nikolay [14]

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Step-by-step explanation:

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3 years ago

A teacher administers three different versions of a test to his students. The two-way table shows the results. Find each probabi

kakasveta [241]

The P(pass | Test A) is 31.81%, P(pass | Test B) is 29.87%, and P(pass | Test C) is 22.07%

<h3>What is probability?</h3>

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

Part A:

Total number of students = 154

Total number of students passed the test = 49

The probability:

P(pass | Test A) = 49/154 = 31.81%

Part B:

Total number of students = 154

Total number of students passed the test = 46

The probability:

P(pass | Test B) = 46/154 = 29.87%

Similarly, for the probability:

P(pass | Test C) = 34/154 = 22.07%

Thus, the P(pass | Test A) is 31.81%, P(pass | Test B) is 29.87%, and P(pass | Test C) is 22.07%

Learn more about the probability here:

brainly.com/question/11234923

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2 years ago

Multiply. Write your answer in simplest form.<br><br> 5/7 x 7/10

grandymaker [24]

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3 years ago

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flight

Bezzdna [24]

Answer:

a. p=1.000

b. p=0.2924

c. p=0.7358

d. No

Step-by-step explanation:

a. This problem satisfies all the criteria for a binomial experiment expressed as:

$P(X=x){n\choose x}p^x(1-p)^{n-x}$

-Given that p=0.85, n=14, the probability that exactly all 14 were on time is calculated as:

$P(X\geq 1)=1-P(X=0)\\\\=1-{12\choose 0}0.85^0(1-0.85)^{12}\\\\=1-1.297\times 10^{-10}\\\\=1.0000$

Hence, the probability that all 12 flights are on time is 1.0000

b. Given that n=12, and p=0.85

-The probability that exactly 10 flights are on time is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924$

Hence, the probability that exactly 10 flights are on time is 0.2924

c. Given that n=12, and p=0.85

-The probability that more of 10 or more flights are on time:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358$

Hence, the probability of 10+ flights being on time is 0.7358

d. We first find the mean of the distribution:

$\mu=E(X)=0.85\times 14\\\\=11.9$

#We then find the probability of 11+=0.3012+0.1422=0.4434

-We compare the expectation to the probability of 11+ flights being on time.

No. Since the probability $P(X\geq 10)$ =0.4434 < that the expectation, 11.9, it is not unusual for 11+ flights to be on time.

*I have used a sample size of n=12 since there are two separate n values:

5 0

3 years ago