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3 years ago

David hikes 2 ¼ miles in ½ an hour at this rate, how long will it take him to walk 18 miles?

1 answer:

5 0

So in 30 min will hike 2 1/4 miles =9 of those 1/4 miles therefore in 10 min will hike 9/3 =3 of those 1/4 miles = 3/4 of a mile

3/4 mile in 10 min;
(3/4)• 4 = 3 miles in 40 min;
6 miles in 80 min;
9 miles in 120 min;
12 miles in 160 min;
15 miles in 200 min;
18 miles in 240 min or 4 hours

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I need help with 15 please ASAP!!!!!!

Lady bird [3.3K]

Answer:

x must equal 5 and y must equal 2

therefore,

5a + 2b = 23 and

5a - 2b = 7.

I'm gonna take 5a + 2b = 23 and add 7 to each side:

5a + 2b + 7 = 30

and substitute 5a -2b in for 7 on the left side:

5a + 2b + (5a-2b) = 30

10a = 30

a = 3

Now we know a. Now take

5a -2b = 7

and substitute in 3 for a

5*(3) - 2b = 7

15 -2b = 7

8 = 2b

b = 4

a = 3 and b = 4

Step-by-step explanation:

6 0

3 years ago

What is the LCM of 16a 3 b and 24ab 2?

vitfil [10]

Do the operation:

3, 16, 24 | 2

3, 8, 12 | 2

3, 4, 6 | 2

3, 2, 3 | 2

3, 1, 3 | 3

1, 1

Answer:

$\boxed{\bf~The~answer~is~2^{4}~*~3~or~48.}$

Hope it helped,

Happy homework/ study/ exam!

6 0

3 years ago

An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$