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Nesterboy [21]
4 years ago
6

How would I do this please help.

Mathematics
1 answer:
Tomtit [17]4 years ago
6 0

Answer:

x=10

1. Set them equal to each other

2. Add ll on each side

3. Subtract x from each side

4. divde 2x from 20 and it leaves you with 10

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Please answer the question carefully and completely.
LiRa [457]

He has 21 classmates, you know this by, taking 17.5 and dividing it by the 5/6. This leading to get you 21.

7 0
4 years ago
Micah invests $5,280 in an account that earns 4.2% interest, compounded monthly.
Olin [163]

Answer:

Hence By End of the year with monthly compounded interest it will have 5522.56 $

Step-by-step explanation:

Given:

Initial investment =5.280 $

Rate of interest =4.2%

To Find:

Amount after the 1 year

Solution:

As the investment follows the rule for compound interest as ,

A=P(1+R)^t

Here A=amount after t years

R= rate of interest , P= principal amount   t is time period

So given is monthly compounded interest

so t will divided into 12 parts as there 12 months in one year.

P=5280 $ , R=4.2/12 %  , t=12

A=5280(1+0.045/12)^(12)

A=5280(1.00375)^12

A=5280(1.046)

A=5522.56 $

8 0
3 years ago
Researchers have found that after 25 years of age the average size of the pupil in a
laila [671]

Answer:

At 25 = 6.8612mm

At 50 years = 5.422mm

Step-by-step explanation:

Equation,

d = 2.115Logₑa + 13.669

d = diameter of the pupil

a = number of years

Note : Logₑa = In a (check logarithmic rule)

d = 2.115Ina + 13.669

1. At 25 years,

d = -2.115In25 + 13.669

d = -2.115 × 3.2188 + 13.669

d = -6.807762 + 13.669

d = 6.8612mm

At 25 years, the pupil shrinks by 6.86mm

2. At 50 years,

d = -2.1158In50 + 13.669

d = -2.1158 * 3.912 + 13.669

d = -8.2770 + 13.699

d = 5.422mm

At 50 years, the pupil shirks by 5.422mm

To save this question, I had to plug in the values into the equation.

Solving for Logₑa might be difficult, so instead I used Inx which is the same thing. Afterwards, i substituted in the values and solve the equation for each years.

8 0
3 years ago
In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp
nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

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

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 189000, \sigma = 20500, n = 50, s = \frac{20500}{\sqrt{50}} = 2899.14

The probability that the sample mean will be $192,000 or more is

This is 1 subtracted by the pvalue of z when X = 192000. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{192000 - 189000}{2899.14}

Z = 1.03

Z = 1.03 has a pvalue of 0.8485.

1-0.8485 = 0.1515

15.15% probability that the sample mean will be $192,000 or more.

7 0
3 years ago
Make 2 equations for this geometric sequence:<br> 36, 18, 9, 4.5, 2.25, 1.125, 0.5625, 0.28125...
NNADVOKAT [17]
I think the answer will be .5
3 0
3 years ago
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