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

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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 $k$ successes 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.