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Margarita [4]
3 years ago
12

Carol jogged 1 3/4 miles on 5 days last week.She jogged 2 1/4 miles on 4 days this week.Was her total distance greater last week

or this week?How much greater? Explain
Mathematics
1 answer:
elena55 [62]3 years ago
6 0

Answer: Her total distance was greater this week.

Step-by-step explanation:

Carol jogged 1 3/4 miles on 5 days last week. Converting 1 3/4 miles to decimal, it becomes 1.75 miles. The total number of miles that she jogged last week is

5 × 1.75 = 8.75 miles

She jogged 2 1/4 miles on 4 days this week. Converting 2 1/4 miles to decimal, it becomes 2.25 miles. The total number of miles that she jogged this week is

4 × 2.25 = 9 miles

Since 9 miles is greater than 8 miles, then her total distance was greater this week.

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What is the perimeter of triangle OJL?<br><br> A.24, B.23, C. 19, D.17
MrRissso [65]

Answer: 24

Step-by-step explanation:

7 0
3 years ago
The average length of a ladybug can range from 0.08 to 0.4 inches. find two lengths that are within the given span. write them a
inysia [295]

<u>Answer</u>

1) 11/100

2) 1/10


<u>Experience</u>

There are infinate fractions between any two integers. The same way there are somany numbers between 0.08 and 0.4.

Examples are 0.09, 0.10, 0.11. 0.12,0.2,0.3, 0.35 and many more.

I will just pick two of them and write them as a fraction.

1) 0.11 = 11/100

2) 0.20 = 20/200

             = 2/20

             = 1/10


3 0
3 years ago
Read 2 more answers
Help please I don’t get it
Licemer1 [7]
The question is asking that given point k and m is such then figure out point L. this shouldn't be the whole question. is there a diagram that comes with it?
7 0
3 years ago
Find the “x” in simplest form
kirill115 [55]

Answer:

x = -7/5  + 1/5  √22

Step-by-step explanation:

Hope this Helps!!

6 0
3 years ago
What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p? C
charle [14.2K]

Answer:

(D)E[ X ] =np.

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success​ p,

f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq  x\leq n

E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}

Now,

x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)

Substituting,

E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}

Factoring out the n and one p from the above expression:

E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}

Representing k=x-1 in the above gives us:

E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}

This can then be written by the Binomial Formula as:

E[ X ] = (np) (p +(1 - p))^{n -1 }= np.

5 0
3 years ago
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