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3 years ago

What is 4 1/2 times(×) 1 1/6

Mathematics

2 answers:

sveticcg [70]3 years ago

7 0

Answer:

5.25

Step-by-step explanation:

Basile [38]3 years ago

3 0

[ Answer ]

$\boxed{5 \frac{1}{4} }$

[ Explanation ]

$4\frac{1}{2}$ · $1\frac{1}{6}$

Turn Mixed Numbers Into Improper Fractions:

$4\frac{1}{2}$ = $\frac{9}{2}$

$1\frac{1}{6}$ = $\frac{7}{6}$

Multiply Across:

$\frac{9*7}{2*6}$

= $\frac{63}{12}$

Simplify:

= $5\frac{1}{4}$

$\boxed{[ \ Eclipsed \ ]}$

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The Census Bureau's Current Population Survey shows that 28% of individuals, ages 25 and older, have completed four years of col

gizmo_the_mogwai [7]

Answer:

19.35% probability that five will have completed four years of college

Step-by-step explanation:

For each individual chosen, there are only two possible outcomes. Either they have completed fourr years of college, or they have not. The probability of an adult completing four years of college is independent of any other adult. So the binomial probability distribution is used 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.

28% of individuals

This means that $p = 0.25$

For a sample of 15 individuals, ages 25 and older, what is the probability that five will have completed four years of college?

This is P(X = 5) when n = 15. So

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

$P(X = 5) = C_{15,5}.(0.28)^{5}.(0.72)^{10} = 0.1935$

19.35% probability that five will have completed four years of college

5 0

3 years ago

What is the answer /6=40/60

ddd [48]

There is no answer, the equation is false.

7 0

3 years ago

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midni

Alex17521 [72]

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, $\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13$

p = 0.20

So, hypothesis:

$H_0:p=\hat{p}\\\\H_a:\hat{p}$

so, test statistic value would be

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604$

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

4 0

3 years ago

Read 2 more answers

Suppose that A = PDP-1 . Prove that det(A) = det(D)

daser333 [38]

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that for two matrices A and B we have that:

det(AB) = det(A)*det(B) and $det(A^{-1}) = \frac{1}{det(A)}$ . Now:

$A = PDP^{-1}$

$det(A) = det(PDP^{-1})$

$det(A) = det(P)*det(D)*det(P^{-1})$

$det(A) = det(P)*det(D)*\frac{1}{det(P)}$

$det(A) = det(P)*\frac{1}{det(P)}*det(D)$

$det(A) = det(D)$ .

5 0

3 years ago

PLZZZZ NEED HELP!!!!!!

jeka94

Notice the 7) 1st sequence, and the 8) second one

$\bf \begin{array}{llll}
term&value\\
-----&-----\\
1&t_1=-2(1)+5\\
2&t_2=-2(2)+5\\
3&t_3=-2(3)+5\\
4&t_4=-2(4)+5\\
5&t_5=-2(5)+5\\
6&t_6=-2(6)+5\\
7&t_7=-2(7)+5\\
8&t_8=-2(8)+5\\
9&t_9=-2(9)+5\\
\end{array}\qquad \qquad 
\begin{array}{llll}
term&value\\
-----&-----\\1&t_1=2(3)^{1-1}\\2&t_2=2(3)^{2-1}\\3&t_3=2(3)^{3-1}\\\end{array}$

7 0

3 years ago