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3 years ago

How do you solve 2y-7y=5

Mathematics

1 answer:

scZoUnD [109]3 years ago

6 0

Answer:

$y=-1$

Step-by-step explanation:

1. Combine like terms

2y-7y = -5y

-5y=5

2. Divide by -5 and the answer is $y=-1$

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lidiya [134]

I'm not going to give you the answer but I'll help you.

Step-by-step explanation:

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3 years ago

. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir

mrs_skeptik [129]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$P(U |D ) = 0.198$

$P(O\ n \ B) = 0.188$

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

Also

$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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3 years ago

Not sure how to solve the problem in the pic :)

yulyashka [42]

Answer:

Step-by-step explanation:

do the division first then subtract 11

x/3 - 11

8 0

3 years ago

Perform the indicated operation.<br> f(n) = 4n - 5, g(n) = n^2 - 5<br> Find: (f - g) (n)

Alex777 [14]

Hi there!

$\large\boxed{-n^2 + 4n}$

Begin by evaluating f - g:

4n - 5 - (n² - 5)

Simplify:

4n - 5 - n² + 5 = 4n - n²

Or it can be written as -n² + 4n to match the first answer choice.

3 0

2 years ago

A human resources representative claims that the proportion of employees earning more than $50,000 is less than 40%. To test thi

bearhunter [10]

Answer:

The statistic for this case would be:

$z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}$

And replacing we got:

$z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92$

Step-by-step explanation:

For this case we have the following info:

$n =700$ represent the sample size

$X= 305$ represent the number of employees that earn more than 50000

$\hat p=\frac{305}{700}= 0.436$

We want to test the following hypothesis:

Nul hyp. $p \leq 0.4$

Alternative hyp : $p>0.4$

The statistic for this case would be:

$z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}$

And replacing we got:

$z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92$

And the p value would be given by:

$p_v = P(z>1.922)= 0.0274$

4 0

3 years ago