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frez [133]
3 years ago
14

Skylar and Rodrigo each recorded how far they traveled while skateboarding. Skylar traveled 65 feet in 5 seconds and Rodrigo tra

veled 108 feet in 8 seconds. How much farther did Rodrigo travel per second than Skylar?
PLEASE PUT YOUR ANSWER IN FEET !
Mathematics
1 answer:
Elena-2011 [213]3 years ago
4 0
Skylar went 13ft per second
Rodrigo went 13 1/2 ft per second
So, 1/2 a ft per second
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The length of a rectangle is represented by x2 + 3x + 2, and the width is represented by 4x.
Amanda [17]
A: Perimeter = 2*length + 2*width = 2x^2 + 14x + 4

B: Area = length*width = 4x^3 + 12x^2 + 8x
5 0
3 years ago
A store manager instructs some new trainees that a leather chair priced at 200$ winds up costing A customer $218 once the sales
Ulleksa [173]

Answer:

9.00%

Step-by-step explanation:

So, we would do

18 divided by 200 = 9.00%

Therefore your answer will be 9.00%

Hope this helps!

5 0
3 years ago
Devon scored 18 points in the first basketball game but 22 in the second game. What’s the percent of change?
fenix001 [56]

Answer:

4%?

Step-by-step explanation:

I honestly really don't know just don't even think about my answer

6 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean \mu =266\\

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find P(X\geq 308)

Using normal distribution,

z=\frac{X-\mu}{\sigma}

z=\frac{308-266}{15}

=\frac{42}{15}

Thus z=2.8

So P\left (X\geq 308  \right )=1-P(X

=1-P(z

=1-Table\:  value\:  of\:  2.8

=1-0.99744

=0.00256

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu

This implies X=-1.8808*15+266=237.788

Thus the babies who are born on or before 238 days are considered to be premature.

5 0
3 years ago
A lock is using 3 digits from 1 to 7 as its lock combination, the numbers can be repeated. How many different possible lock comb
Oksi-84 [34.3K]

The question is an illustration of selection (or combination)

The number of lock combination is 343

The given parameters are:

<em />n = 7<em> --- the number of digits to select from</em>

<em />r = 3<em> --- the number of digits the lock uses</em>

<em />

Because the digits can be repeated, the number of lock combinations is:

Lock = n^r

So, we have:

Lock = 7^3

Lock = 343

Hence, the number of lock combination is 343

Read more about combinations at:

brainly.com/question/15301090

7 0
3 years ago
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