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Burka [1]
3 years ago
6

Jenny drove 169 miles in 3 hours and 30 minutes how many is it in miles per minute

Mathematics
2 answers:
ddd [48]3 years ago
7 0
Convert 3 hours and 30 minutes all into minutes so that would be 210 minutes. You then have 169 miles / 210 minutes. Get the denominator to 1 minutes by dividing by 210 to the top and bottom and you'll get .080476 miles / 1 minute which is the answer.
sweet [91]3 years ago
3 0

There is 60 minutes in an hour so 210minutes

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9. A ranch has 575 cows on 25 acres of land. Which rate best represents the relationship
Vitek1552 [10]

Answer:

10/10

Step-by-step explanation:

7 0
3 years ago
Solve for x. *<br> (5x + 13)<br> (9x - 31)
ycow [4]

Answer:

x=11

Step-by-step explanation:

5x+13=9x-31

subtract 9 from both sides.

-4x+13=-31

you want to isolate the x value so subtract 13 from both sides.

-4x=-44

now divide both sides by -4.

x=11

6 0
3 years ago
Solve i + j + t (3i - j) = -i + s (j)<br> In the context of Vector Equations of a Line
Oksana_A [137]

Answer:

r = i + j + (-2/3)(3i - j)

Step-by-step explanation:

Vector Equation of a line - r = a + kb ; where r is the resultant vector of adding vector a and vector b and k is a constant

if a = i + j ; b = t(3i - j) and r = -i +s(j)

for this to be true all the vector components must be equal

summing i 's

i + 3ti  = -i; then t = -2/3

j - tj = sj; then s = 1-t; substitue t; s=1+2/3 = 5/3

so r = i + j + (-2/3)(3i - j) which will symplify to -i + 5/3j

3 0
2 years ago
Find m &lt;pml ?????????
Gnesinka [82]
Angle pml =39
Hope this helps u....
5 0
2 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
2 years ago
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