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2 years ago

If u get 2 points every 5 minutes, how long will it take to get 1,000 points?

Mathematics

2 answers:

yulyashka [42]2 years ago

6 0

2 points : 5 minutes

2 x 500 points : 5 x 500 minutes

1000 points : 2500 minutes

It will take 2500 minutes to get 1000 points.

marshall27 [118]2 years ago

3 0

Answer:
2500 Minutes

Step-by-step explanation:

If you get 2 points every 5 minutes, how long will it take it to get 1000 points? It's simple

Divide the Number of Points by the Number Of Points you get every 5 minutes then times it by the number of minutes you get 2 points making 2,500 Minutes

I hope it helped ya!

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What is the value of x?

Helen [10]

if the original value of x was negative then X the opposite sign version of X would have to be positive Princeton's if I start with x equals -3 then x equals -3 equals + 3 which is positive

8 0

2 years ago

Find the probability p(e or f) if e and f are mutually exclusive, p(e)equals 0.25, and p(f)equals 0.51.

Masteriza [31]

The definition of two events being mutually exclusive (or disjoint) only means that it is not possible for the two events to occur together. Given two events, E and F, they are mutually exclusive and also mean independent.

In this case, since events E and F are mutually exclusive, therefore the probability that either E or F will occur will simply be the sum of two events.

P (E or F) = P (E) + P (F)

P (E or F) = 0.25 + 0.51

P (E or F) = 0.76

Therefore this means that there is a 76% probability that either E or F will occur.

8 0

3 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

<em>Hope this helps!</em>

7 0

2 years ago

Plz plz plz plz help plz no LINKS

Viefleur [7K]

(2.7, 2.3)
(2, 1.7)
(1.7, 1.3)
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5 0

3 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago