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3 years ago

Decide whether the two inequalities are equivalent. Explain.

Mathematics

1 answer:

zhenek [66]3 years ago

6 0

Answer:

D,no if you graph both inequalitiesvon thevsame geaph ,you will get b =16

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Use long division to find 5210 divided by 46

In-s [12.5K]

52-46=6 drop down the 1 = 61- 46=15 drop down the 0 =150 46x3= 138 so 150-138= 12

3 0

4 years ago

Can you help me with number 10

rewona [7]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(3^{-2})^2}{3^{-3}}\implies \cfrac{3^{-2\cdot 2}}{3^{-3}}\implies \cfrac{3^{-4}}{3^{-3}}\implies \cfrac{1}{3^{-3}\cdot 3^4}\implies \cfrac{1}{3^{-3+4}}\implies \cfrac{1}{3^1}\implies \cfrac{1}{3}$

4 0

4 years ago

I need to know what the answer is the question is Josie bikes 40 miles each months for five months she multiplies 40×5 what unit

sleet_krkn [62]

The product would be 200 obliviously it would be miles because it says five months so it’s implying that you need to find the amount biked in 5 months

So it’s should be miles

3 0

3 years ago

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

lubasha [3.4K]

Answer:

a. 10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. 0.2457 = 24.57% probability 10 of the problems can be resolved today

c. 0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today

d. 0.5017 = 50.17% probability more than 10 of the problems can be resolved today

Step-by-step explanation:

For each case, there are only two possible outcomes. Either they are solved, or they are not. Cases are independent of each other. 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

It can resolve customer problems the same day they are reported in 80% of the cases.

This means that $p = 0.8$

Sample of 13 cases:

This means that $n = 13$

a. How many of the problems would you expect to be resolved today? What is the standard deviation?

Expected:

$E(X) = np = 13*0.8 = 10.4$

Standard deviation:

$\sqrt{V(X)} = \sqrt{13*0.8*0.2} = 1.44$

10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. What is the probability 10 of the problems can be resolved today?

This is $P(X = 10)$ . So

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

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

0.2457 = 24.57% probability 10 of the problems can be resolved today.

c. What is the probability 10 or 11 of the problems can be resolved today?

This is $p = P(X = 10) + P(X = 11)$ . So

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

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$p = P(X = 10) + P(X = 11) = 0.2457 + 0.2680 = 0.5137$

0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today.

d. What is the probability more than 10 of the problems can be resolved today?

This is

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13)$ . So

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

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$P(X = 12) = C_{13,12}.(0.8)^{12}.(0.2)^{1} = 0.1787$

$P(X = 13) = C_{13,13}.(0.8)^{13}.(0.2)^{0} = 0.0550$

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) = 0.2680 + 0.1787 + 0.0550 = 0.5017$

0.5017 = 50.17% probability more than 10 of the problems can be resolved today

6 0

4 years ago

Make each expression a true statement

Lelu [443]

Answer:

( d / c ) < ( a / d )

( b / d ) < ( a / d )

( d / a ) < ( a / d )

Step-by-step explanation:

For this problem, we need to consider the given information that a > b > c > d > 0. Note that a is the greatest number and that d is the least number.

Given this fact, we can create a fraction that will result in a maximum value by simply doing the following:

a / d

The reason the fraction ( a / d ) results in a maximum, is because of the given information where a is the furthest from 0, and d is the closest to 0. Therefore, we can say that no other combination will result in a value greater than ( a / d).

So in the three given statements, we have different combinations of fractions being compared to our maximum value fraction. Our maximum value fraction ( a / d ) will always be greater than any other combination.

Thus, we will choose the sign that makes ( a / d ) greater than the other choice fractions.

Cheers.

6 0

3 years ago