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maks197457 [2]
3 years ago
9

A woman deposits 100 EUR in her daughter's bank account on her first birthday. On every subsequent birthday, she deposits 10 EUR

more than she deposited the previous year, so on her second birthday, she deposits 110 EUR, and on her third birthday she deposits 120 EUR.
By the time her daughter is 21 years old, how much money has been deposited in her account?
Mathematics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

By the time her daughter is 21 years old 300 EUR will have been deposited into the account.

Step-by-step explanation:

Since a woman deposits 100 EUR in her daughter's bank account on her first birthday, and on every subsequent birthday, she deposits 10 EUR more than she deposited the previous year, so on her second birthday, she deposits 110 EUR, and on her third birthday she deposits 120 EUR, to determine, by the time her daughter is 21 years old, how much money has been deposited in her account, the following calculation must be performed:

100 + (20 x 10) = X

100 + 200 = X

300 = X

Therefore, by the time her daughter is 21 years old 300 EUR will have been deposited into the account.

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According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard devi
Mekhanik [1.2K]

Answer:

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is mean amount of inches of rain and \sigma is the standard deviation.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Mean \mu, standard deviation \sigma

n days:

This means that s = \frac{\sigma}{\sqrt{n}}

Applying the Central Limit Theorem to the z-score formula.

Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is mean amount of inches of rain and \sigma is the standard deviation.

5 0
3 years ago
On Sunday, Sheldon bought 4 1/2 kg of plant food. He used 1 2/3 kg on his strawberry plants and used 1/4 kg for his tomato plant
VladimirAG [237]

Answer:

a) 2\frac{7}{12} kg of plant food did Sheldon have left.

b) She need is \frac{43}{12} kg and she don't have enough to left.

Step-by-step explanation:

Given : On Sunday, Sheldon bought 4\frac{1}{2} kg of plant food. He used 1\frac{2}{3} kg on his strawberry plants and used \frac{1}{4} kg for his tomato plants.

To find :

a) How many kilograms of plant food did Sheldon have left? Write one or more equations to show how you reached your answer.

Solution :

Sheldon bought plant food 4\frac{1}{2}=\frac{9}{2} kg.

He used strawberry plants 1\frac{2}{3}=\frac{5}{3} kg.

He used tomato plants \frac{1}{4} kg

Kilograms of plant food did Sheldon have left is given by,

L=\frac{9}{2}-(\frac{5}{3}+\frac{1}{4})

L=\frac{9}{2}-\frac{23}{12}

L=\frac{54-23}{12}

L=\frac{31}{12}

L=2\frac{7}{12}

Therefore, 2\frac{7}{12} kg of plant food did Sheldon have left.

b) Sheldon wants to feed his strawberry plants 2 more times and his tomato plants one more time. He will use the same amounts of plant food as before. How much plant food will he need? Does he have enough left to do so?

Solution :

Sheldon wants to feed his strawberry plants 2 more times.

i.e. Strawberry plants used is 2(\frac{5}{3})=\frac{10}{3} kg.

His tomato plants one more time i.e. \frac{1}{4} kg.

Total plant she need is \frac{10}{3}+\frac{1}{4}=\frac{43}{12}

As \frac{31}{12}

No she don't have enough to left.

4 0
3 years ago
- 12 = 2 + 5d + 2d <br> ................
zmey [24]

Answer:

-2

Step-by-step explanation:

12=2+7d

7d= -12-2

7d= -14

d= -14

-----

7

d= -2

4 0
3 years ago
Read 2 more answers
2p/3 - 12 = -2<br> A: 15<br> B: 7<br> C: 17<br> D: -21
Naddik [55]

the answer is A because indont know 9bp

3 0
3 years ago
Graph a system of equations to approximate the
SIZIF [17.4K]

Answer: B

Step-by-step explanation:

8 0
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