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IRINA_888 [86]
2 years ago
14

In June, Meg runs a race 10 seconds faster than she did in April. Let y represent her finishing time for the race in April. Whic

h expression represents her faster time in June?
y - 10
y + 10
10 - y
Mathematics
2 answers:
Sophie [7]2 years ago
8 0

Answer:

y+10

Step-by-step explanation:

zavuch27 [327]2 years ago
8 0

Answer:

The first one

Step-by-step explanation:

Y-10 bc she ran 10 seconds faster

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sleet_krkn [62]
For this question, you need to find the price per unit. The unit in this case is a note, so we are looking for the price of each individual note in the package. The way we find this is by taking the price divided by the number of notes So package one would come out to be (9/18) or .50 dollars per note. Package two would come out to be (12/16) or .75 dollars per not. Package three would be (4/10) or .40 dollars per note. Your answer would simply the one with the lowest price per note, and in this case would be package three
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3 years ago
PLEASE HELP I WILL GIVE A LOT OF POINTS
FromTheMoon [43]

Answer:

765 ft

Step-by-step explanation:

1/2(30 times 22)=330 times 2= 660+ 1/2(21 times 10)=105

3 0
3 years ago
Read 2 more answers
Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.
MrMuchimi
A random variable following a binomial distribution over n trials with success probability p has PMF

f_X(x)=\dbinom nxp^x(1-p)^{n-x}

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1

The mean is given by the expected value of the distribution,

\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}

The remaining sum has a summand which is the PMF of yet another binomial distribution with n-1 trials and the same success probability, so the sum is 1 and you're left with

\mathbb E(x)=np=126\times0.27=34.02

You can similarly derive the variance by computing \mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2, but I'll leave that as an exercise for you. You would find that \mathbb V(X)=np(1-p), so the variance here would be

\mathbb V(X)=125\times0.27\times0.73=24.8346

The standard deviation is just the square root of the variance, which is

\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834
7 0
3 years ago
Through:(-1, 2) and (0,2)
Elis [28]
What are you asking for exactly ?
5 0
2 years ago
18 Ib. Then her dog loses another 1 lb. Finally, it
Pavel [41]

Answer:

17 lb I guess 8 don't think the question is full

Step-by-step explanation:

the question is half

8 0
3 years ago
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