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3 years ago

14

oel biked 9 miles east and then 12 miles north. If he biked back to his starting point using the most direct route, how many mil

es would he ride all together?

1 answer:

kvasek [131]3 years ago

6 0

Answer:

36 miles

Step-by-step explanation:

(Imagine a huge triangle on a perfectly open space-- not with streets or houses or cornfields.)

The dimensions of the two sides given are multiples of a classic Pythagorean Triple: 3, 4, 5. $3^{2} + 4^{2} = 5^{2}$

9 is 3x3, 12 is 3x4 so the remaining side would be 3x5 15

Add those 3 sides to get the total distance, 36 miles.

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The volume of a sphere is 4,500π cubic yards. What is the radius of the sphere?

algol [13]

Answer:

15

Step-by-step explanation:

In the question it is given that the volume of a sphere is 4,500π cubic yards. And we have to find the radius.

We know that,

4/3πr³ = Volume of a sphere

Now, Substituting the values in the fomulae :

⇒ 4/3πr³ = 4500π

Cancelling π from both sides :

⇒ 4/3r³ = 4500

⇒ 4r³ = 4500 × 3

⇒ 4r³ = 13500

⇒ r³ = 13500/4

⇒ r³ = 3375

Taking cube root on both sides we get :

⇒ r = 15

Therefore,

<u>The radius of the sphere is 15 yards</u>

5 0

2 years ago

Read 2 more answers

After six biology tests, Ruben has a mean score of 74. What score does Ruben need on the next test to raise his average (mean) t

blondinia [14]

Answer:

88

Step-by-step explanation:

this question is a question that involves the use of the statistical principles of finding average

let the summation of the six scores of the biology test be represented as ∑x

∴ ∑x÷ 6=74

∑x=74 × 6

=444

444 is the summation of all the scores of all the six biology scores reuben had

the next test would be his 7th

let y represent the score for the 7th test

therefore, $\frac{444 + y}{7}$ = 76

444 + y = (76 × 7)

444 + y = 532

y=88

Reuben needs to score 88 to raise his average to 76

4 0

2 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

So

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago

I NEED THE ANSWER ASAP PLEASE HELP!!

maksim [4K]

Answer:C

Explanation:none

8 0

2 years ago

62 is 50% of what number?

Stels [109]

62 is 50% of 124 (62*2=124)

5 0

3 years ago

Read 2 more answers