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3 years ago

Solve for m -mk-90>85

1 answer:

5 0

First add 90 to both sides) -mk>85+90simplify 85+90 to 175) -mk>175divide both sides by K) -m>175/kmultiply both sides by -1) m<-175/k
YOUR ANSWER IS
M<-175/K
hope l helped today l was stuck on the same question doing part 2 of alg exam , and l came through your question l hope it's not too late :)

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Please reference attached image for the problem that requires solving. Thank you so much for taking the time to help.

Lemur [1.5K]

Explanation:

The number of times the 6-sided number cube will be rolled will is

$750$

Let the numbers greater than 4 be represented below as

$E_1$ $\begin{gathered} E_1=\lbrace5,6\rbrace \\ n(E_1)=2 \end{gathered}$

The number of sample space will be

$n(S)=6$

The probability of rolling a number greater than 4 will be calculated below as

$\begin{gathered} Pr(E_1)=\frac{n(E_1)}{n(S)} \\ Pr(E_1)=\frac{2}{6}=\frac{1}{3} \end{gathered}$

Hence,

To calculate the number of times a number greater than 4 will be rolled will be calculated below as

$\begin{gathered} =Pr(E_1)\times750 \\ =\frac{1}{3}\times750 \\ =250times \end{gathered}$

Hence,

The final answer is

$\Rightarrow250\text{ }times$

5 0

1 year ago

Describe the relationship betw

GalinKa [24]

Answer:

108, 324, 972

Step-by-step explanation:

4 (*3)

ans. (*3)

4 0

3 years ago

Read 2 more answers

= 3when x- What is the value for f(x) = 3^2x+1 when x=1/2 3 6 9 15

Zigmanuir [339]

Its 9 because it includes f(x)

6 0

3 years ago

What is the lcm of 8,12,3

kramer

Answer: 24

Step-by-step explanation:

Lowest common multiple means the lowest number that can be divisible by 8,12,3

Looking at this, let's pick 8&3 first

8×3=24

12×3=36

12×8=96

Looking at this value,24 seems to be the lowest,

Therefore the lowest common multiple is 24

5 0

3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago