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3 years ago

47 is the result when 12 and x are added. What is the value of x? Enter your answer in the box as a fraction in simplest form.

Mathematics

2 answers:

seraphim [82]3 years ago

6 0

The correct value for x is 3.9

aleksklad [387]3 years ago

5 0

Answer:

x=1/14

don't know how the other person got such a wild answer

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-3|x|+2x-1 if x=-5<br><br> plz help

Savatey [412]

Answer:

f(x) = -3|x| + 2x - 1

f(-5) = -3|-5| + 2*(-5) - 1

= -3*5 + 2*(-5) - 1

= -15 - 10 - 1

= -26

Hope this helps!

5 0

2 years ago

Read 2 more answers

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

f(x) increases by 2 as x increases by 1 x value.

We know it isn't A since and x value plue 2 would not equal the y value.

But if you multiply each x-value by 2 you get f(x)

8 0

2 years ago

Read 2 more answers

Help please with the process of solving

jonny [76]

Answer:

Slope (m) =ΔY /ΔX =-3/ 1 =-3

n slope = -3

From equation [<u>-2</u>x-8]

m slope = -2

so B) m>n

Step-by-step explanation:

5 0

3 years ago

How would I explain the slope and y-intercept of the equation y=8x+10???? need help soon please and thanks.

Dimas [21]

Step-by-step explanation:

with the formula y= mx+c the slope is m and the y intercept is c so the slope of this equation is 8 and the y intercept is 10

6 0

2 years ago

One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will ap

pashok25 [27]

Answer:

0.19

Step-by-step explanation:

We are given that

Number of rolls of die=n=1000

Let the event of six coming up be success.Then, in each trial , the probability of success =p=P(success)=P(6)= $\frac{1}{6}$

Let X be the random variable for the number of sixes in the 1000 rolls of die.

Then, $X\sim Binom(1000,\frac{1}{6})$

Since, n is very large,the binomial random variable can be approximated as normal random variable.

Mean, $\mu=np=1000\times \frac{1}{6}=166.67$

Variance= $\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89$

$X\sim N(166.67,138.89)$

$P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]$

= $P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z$

= $\phi(2.83)-\phi(-1.41)$

=0.9977-0.0793=0.9184

Thus, the probability that the number 6 appears between 150 to 200 times=0.92

Now, given that 6 appears exactly 200 times .

Therefore, other number appear in other 800 rolls .

We have to find the probability that the number 5 will appear less than 150 times.

Therefore, for 800 rolls, let the event of 5 coming up be success.

Then , p=P(success)=P(5)= $\frac{1}{5}$

Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.

Then, $Y\sim bin(800,\frac{1}{5})$

Mean, $\mu=np=800\times \frac{1}{5}=160$

Variance, $\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128$

$Y\sim N(160,128)$ because n is large

$P(Y$

$P(Y$

Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19

4 0

2 years ago