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3 years ago

Help me its due today

Mathematics

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

$65.35

Step-by-step explanation:

First you have to divide 5.4% by 100
Take 5.4/100 and multiply it by 62
Take the number you get from step 2 and add it to 62! (:

You might be interested in

Can you please tell me how to solve this?

Sedaia [141]

The answer is 142°

Explanation:

angle 1 and angle 2 form are supplementary.

m∠1 + m∠2 = 180°

38° + m∠2 = 180°

m∠2 = 180° - 38°

m∠2 = 142°

Hope it helps!

5 0

2 years ago

Mary who is 13 years old wants to have $8500 to travel to Argentina when she’s 21 years old. She currently has $6439 in savings

marishachu [46]

Answer:

a; she will have $8812

b: It will be enough for her trip

Step-by-step explanation:

In this question, we are tasked with calculating how much a certain value in a savings account that is earning an interest that is compounded annually will be worth.

To calculate this, we use the compound interest formula;

A = P( $(1+r/n)^{nt}$

Where A is the amount after that number of years which of course we want to calculate

P is the principal amount which is the amount we are investing which is $6439 according to the question

r is the interest rate which is 4% = 4/100 = 0.04

t is the time which is 8 years

n is 1 which is the number of times interest will be compounded annually

We plug these values as follows;

A = 6439(1 + 0.04/1)^8

A = 6439(1.04)^8

A = $8,812.22

This amount is greater then the needed $8,500 for the trip and of course it will be enough

8 0

3 years ago

Answer = 615.44 `\('-')/`

babymother [125]

Answer:

answer to what? free points?

Step-by-step explanation:

7 0

2 years ago

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

3 0

3 years ago

Which are the correct measures for

-BARSIC- [3]

Answer:

option c is the correct answer

Step-by-step explanation:

∠YXZ+117°=180°

∠YXZ=180°-117°

∠YXZ=63°

63°+36°+∠YZX=180°(sum of angles is 180)

99°+∠YZX=180

∠YZX=180-99

∠YZX=81

5 0

3 years ago