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bazaltina [42]
3 years ago
5

I NEED HELP ON THIS SO BAD WILL GIVE CROWN

Mathematics
1 answer:
Stells [14]3 years ago
8 0
5.5 quarts equals 22 cups. (There are 4cups in a quart, so 4 * 5.5 = 22)

22-4= 18 cups
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assume that when adults with smartphones are randomly selected, 44% use them in meetings or classes. If 10 adult smartphones use
tensa zangetsu [6.8K]

Solution: The given random experiment follows Binomial distribution with n=10,p=0.44

Let X be the number of adults who use their smartphones in meetings or classes.

Therefore, we have to find:

P(X

We know the binomial model is:

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

\therefore P(X

                       =\binom{10}{0}0.44^{0}(1-0.44)^{12-0}+\binom{10}{1}0.44^{1}(1-0.44)^{10-1}+\binom{10}{2}0.44^{2}(1-0.44)^{10-2}

                       =1 \times 1 \times 0.0030 + 10 \times 0.44 \times 0.0054 + 45 \times 0.1936 \times 0.009672

                       =0.0030+0.0238+0.0843

                       =0.1111

Therefore, the probability that fewer than 3 of them is 0.1111

7 0
3 years ago
If 5|x − 1| = 20, then x = <br> 1 or 5<br> 6 or –8<br> –2 or 4<br> 5 or –3<br> Please help
anygoal [31]

Answer:

Step-by-step explanation:

5 0
3 years ago
Bradley bought 14 yellow highlighters. each highlighter cost $0.94.how much did Bradley spend?
Svetradugi [14.3K]
The answer would be 13.16$

0.94x14=13.16
6 0
3 years ago
A study indicates that 62% of students have have a laptop. You randomly sample 8 students. Find the probability that between 4 a
Scrat [10]

Answer:

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a laptop, or they do not. The probability of a student having a laptop is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A study indicates that 62% of students have have a laptop.

This means that n = 0.62

You randomly sample 8 students.

This means that n = 8

Find the probability that between 4 and 6 (including endpoints) have a laptop.

P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157

P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815

P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297

P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

3 0
3 years ago
On a school playground, a math teacher draws a number line with values from 10 to 10. The distance between each
german

Answer:

12

<h3>Step-by-step explanation:</h3>

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

count the spaces between the two numbers

<h2 />
4 0
3 years ago
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