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Svetllana [295]
3 years ago
10

A multiple-choice test has 7 questions, with 4 possible answers for each

Mathematics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

16384

Step-by-step explanation:

4×4×4×4×4×4×4

= 4⁷

= 16384 ways

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I need help with this . short and easy response please
Alika [10]
\frac{\frac{1}{4}-\frac{2}{3}}{\frac{3}{5}}=\frac{\frac{3}{12}-\frac{8}{12}}{\frac{3}{5}}=\text{ }\frac{-\frac{5}{12}}{\frac{3}{5}}=\text{ -}\frac{25}{36}

7 0
1 year ago
Which expression is the simplest form of (27x^-9)^1/3?
lana [24]
\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\sqrt[n]{a}=a^\frac{1}{n}\\\\(a^n)^m=a^{nm}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(27x^{-9})^\frac{1}{3}=27^\frac{1}{3}(x^{-9})^\frac{1}{3}=\sqrt[3]{27}\cdot x^{-9\cdot\frac{1}{3}}=3x^{-3}=\dfrac{3}{x^3}
8 0
4 years ago
Find z such that the proportion of observations that are less than z in a standard normal distribution is 0.23. (Enter your answ
-BARSIC- [3]

Answer:

-0.74

Step-by-step explanation:

The computation of Z is shown below:

According to the question, data provided in the question is as follows

P(Z >  Z) = 0.23

Based on the above information, the calculation for Z is

Use the table of the standard normal distribution for z

By using this the Z value comes is -0.74

hence, in case of the standard normal distribution, the value of z is -0.74 and the same is to be considered

8 0
4 years ago
Thirty percent of the CFA candidates have a degree in economics. A random sample of three CFA candidates is selected. What is th
Komok [63]

This is an example of a random variable that follows Bernoulli distribution.

Whenever you perform n experiments, with probability p of success, the provability of havin ksuccesses is

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

In this case, you have n=3 (you choose a sample of 3 candidates), p=0.3 (30% of candidates have a degree in economics) and we want k to be at least one.

One way to solve this is to consider that

P(X\geq 1)=P(X=1)+P(X=2)+P(X=3)

But it is much quicker to observe that

P(X\geq 1)=1-P(X

And we have

P(X=0)=\displaystyle\binom{3}{0}0.3^0\cdot 0.7^{3}=1\cdot 1\cdot 0.7^3

So, the probability that at least one of them has a degree in economics is

1-0.7^3 = 0.657

4 0
3 years ago
Don't quite understand lol, help would be appreciated
olga nikolaevna [1]

Answer:

-1/3

Step-by-step explanation:

5 0
3 years ago
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