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2 years ago

15

to train for a race, you plan to run 1 mile the first week and double the number of miles each week for five weeks. How many mil

es will you run for the 5th week. math problem

1 answer:

SashulF [63]2 years ago

3 0

Answer:

16 Miles

Step-by-step explanation:

For every week you simply multiply the number of miles from the previous week by 2, therefore

Week 1: 1

Week 2: 2

Week 3: 4

Week 4: 8

Week 5: 16

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A merchant bought 1000 electric bulbs at the rate of Rs.150 each. 50 of them were broken on the way and the remaining he sold at

neonofarm [45]

Answer:

₹165.79

Step-by-step explanation:

Given:-

No. of electric bulbs = 1000

cost of each electric bulb = ₹ 150

No. of bulbs broken = 50

Selling price of each bulb = x

Profit percentage = 5%

To Find:-

The selling price of each bulb.

Solution:-

Cost price of 1000 electric bulbs,

= 1000 × ₹150

= ₹1,50,000

5% profit on the total cost price,

= {5}/{100}× ₹150000

= ₹7500

Total selling price = ₹157500

No. of bulbs remaining = 950

Therefore, selling price of each bulb,

= {₹157500}/{950}

= ₹165.79

Therefore,

Selling price of each bulb = ₹165.79

5 0

1 year ago

A signal can be sent from one location to another by running different colored flags up a flagpole, one above the other. There a

Verdich [7]

Using the permutation formula, it is found that there are 17,297,280 different signals consisting of 8 flags.

In this problem, the order in which the flags are visited is important, hence the <em>permutation formula</em> is used to solve this question.

<h3>What is the permutation formula?</h3>

The number of possible permutations of x elements from a set of n elements is given by:

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

The first flag is blue, then the remaining 7 are taken from a set of 14, hence:

$P_{(14,7)} = \frac{14!}{7!} = 17,297,280$

There are 17,297,280 different signals consisting of 8 flags.

More can be learned about the permutation formula at brainly.com/question/25925367

8 0

2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

The center of a circle is (4, 6), and an endpoint of a diameter is (2, 5). What is the other endpoint of the diameter?

valentinak56 [21]

Consider this option:
1. if the point (4;6) is the centre of the circle and the point (2;5) is the first endpoint of its diameter, then point (4;6) is the middle point of the diameter (it means that is the middle between the 1st and the 2d endpoints of diameter).
2. using the property described above:
for x of the 2d endpoint of the diameter: x=4*2-2=6;
for y of the 2d endpoint of the diameter: y=6*2-5=7.

answer: (6;7)

3 0

3 years ago

Read 2 more answers

-3x = 36 X = ________ can someone answer this plz

koban [17]

Answer:

-12x

Step-by-step explanation:

36÷(-3)= -12

3 0

3 years ago

Read 2 more answers