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Zolol [24]
3 years ago
15

Megan rides a bicycle 19 miles every 4 hours. At this rate, how many hours did Megan ride if she rode 57 miles?

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0
Megan rode her bicycle for 12 hours.
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Choose the correct equivalent factored form for the expression below.<br> X^4 – 121
VikaD [51]

Answer:

(x^2+11)(x^2-11)

Step-by-step explanation:

You didn't list the options, but it should be:

(x^2 + 11) (x^2 - 11)

We use this formula: a^2 - b^2 = (a + b) (a - b)

8 0
3 years ago
How do you put this question into an equation. *PLEASE ANSWER ASAP*
KatRina [158]
The answer is 0.25t+30m-59.50
4 0
3 years ago
Read 2 more answers
F(x)+g(x)=(x^2+10x+25)+(5-4x)
yKpoI14uk [10]
Just combine like terms: (f+g)(x)=x^2+6x+30
6 0
4 years ago
The Smith family has three children: Tom, John, and Andy. Tom is younger than Andy. John is not older than Tom. From youngest to
blagie [28]

Here, we have to analyse the words and ages

The first statement is "Tom is younger than Andy" means Andy is older than Tom. So first comes Andy and then Tom. Let's Rank Andy =1 and Tom = 2 with 1 assigned to the older.

Now, the next statement is "John is not older than Tom" This means John is younger than Tom. This also means Tom is older than John. So John would become the youngest and hence we assign John = 3 (meaning the youngest)

Ordering from Youngest to oldest would be John (Rank 3), Tom (Rank 2) and Andy (Rank 1)

To conclude the order from youngest to oldest would be John,Tom,Andy

5 0
3 years ago
Read 2 more answers
A lot of 200 chips contain 5 that are defective. 8 are selected random with out replacement.
vaieri [72.5K]

<u>Answer:</u>

1 , \frac{1}{66}  (Answer)

<u>Step-by-step explanation:</u>

Since, 8 chips are selected at random and 5 are there defectives in the lot, So, at least (8 - 5) = 3 chosen chips will be non - defective.

So, P ( At least two of selected part is non defective ) = 1 .

P(first and second samples are defective)

= \frac{5 \times 4}{200 \times 199}

P ( first, second and third samples are defective)

= \frac{5 \times 4 \times 3}{200 \times 199 \times 198}

So,

\frac{{\textrm{Third sample is defective | first and second samples are defectives}}}

=\frac{{\textrm{  P ( first, second and third samples are defective)}}}{{\textrm{  P(first and second samples are defective) }}}

= \frac{3}{198}

= \frac{1}{66}  (Answer)                                                          

7 0
4 years ago
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