The question is incomplete:
When Emily arrived home from school she spent a minutes eating a snack. Then, she spent 15 minutes playing with her dog and 32 doing her chores
. Finally, Emily spent 45 minutes finishing her homework. How long had Emily been home when she finished?
Answer:
Emily had been home for 1 hour and 33 minutes.
Step-by-step explanation:
In order to find the answer, first you have to add up the time she spent in each activity to find the amount of minutes she had been at home when she finished:
1+15+32+45=93
Now, you can express the 93 minutes in hours considering that one hour has 60 minutes. According to this, Emily had been home for 1 hour and 33 minutes (93-60=33).
Answer:
i) P(X<33) = 0.9232
ii) P(X>26) = 0.001
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 30
Given that the standard deviation of the Population = 4
Let 'X' be the Normal distribution
<u>Step(ii):-</u>
i)
Given that the random variable X = 33

>0
P(X<33) = P( Z<1.5)
= 1- P(Z>1.5)
= 1 - ( 0.5 - A(1.5))
= 0.5 + 0.4232
P(X<33) = 0.9232
<u>Step(iii) :-</u>
Given that the random variable X = 26

>0
P(X>26) = P( Z>3.5)
= 0.5 - A(3.5)
= 0.5 - 0.4990
= 0.001
P(X>26) = 0.001
This should have been worth more points, but anyways, here are the answers. Please give thanks :)
1.)Y = 5 - 3X<span>
2.) Y= </span><span>3X - 4</span><span>
3.) Y = 7 - X
4.) Y = -5X
5.) Y = </span>X +6 <span>
6.) Y = 5 -3X
7.) Y = 2 +</span>

<span>X
8.) </span>Y =

X - 3<span>
9.) Y= 2 - [</span>tex] \frac{2}{3} [/tex]<span>
10.) Y = 1 - 5X
11.) Y = 2 - </span>

<span>X
12.) Y = -5 -2X
13.) Y = X - 6
14.) Y = </span>

<span> -2X
15.) Y = 2 +</span>

<span>X
16.) Y = 2 +</span>

<span>X
17.) Y = 1 - 5X
18.) Y = </span>[<span>tex] \frac{3}{4} [/tex]</span>X + 2
<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
The points (0, 3) and (4, -1) are on the graph of y = -x + 3.
I have a site that I use a lot that is perfect for problems like these. It's called Desmos.com. It's basically a free online graphing calculator that is super helpful for most problems relating to graphs.