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2 years ago

If a+1/a = 5, find the value of a² +1/a²

Mathematics

2 answers:

Orlov [11]2 years ago

6 0

a - 1/a =5

take squaring both side

(a - 1/a )^2 = (5)^2

a^2 +1/a^2 -2.a.1/a =25

a^2+1/a^2 -2 =25

a^2+1/a^2 =27 (answer)

docker41 [41]2 years ago

5 0

Answer:

$a^2+\frac{1}{a^2}=23$

Step-by-step explanation:

$a+\frac{1}{a}=5\\squaring\\a^2+(\frac{1}{a})^2+2\times a \times \frac{1}{a}=5^2\\a^2+\frac{1}{a^2} +2=25\\a^2+\frac{1}{a^2}=25-2=23$

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How do you find x? and how do you round to the nearest tenth?

expeople1 [14]

Hey there!

In order to solve this problem, we can use the Pythagorean Theorem, which is $c^2 = a^2 + b^2$ . Since the Pythagorean Theorem would find only half of the length of x, we'll need to double it in the end.

First, let's plug in what we know to the Pythagorean Theorem: $2.1^2 = 1.4^2 + b^2$

Next, we'll simplify to that $b^2$ is by itself:

$b^2 = 2.45$

Then, we'll find the square root of 2.45, multiply that by 2 and round to the nearest tenth:

$2(\sqrt{2.45} ) = 3.1$

Therefore, x = 3.1 ft.

How to round to the nearest tenth:

In a decimal, the tenth is the first number after the decimal point. To round to the tenth, if the number in the hundredth place is less than 5, you keep the number as it is, but if the number in the hundredth place is more than 5, you increase the number in the tenth place by 1.

Hope this helps!! :)

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2 years ago

2.1 & 2.2 Assessment: Vertex & Standard Form of Quadratic Functions

Brums [2.3K]

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4 months ago

I need to know this ASAP

sashaice [31]

Answer:

Problem 1

x=2

y=3

Problem 2

x=-3

y=8

Goodluck :)

3 0

2 years ago

2 Probability and Stats questions!!

mart [117]

1) To find the confidence interval
the sample mean x = 38 σ = 9; n = 85;
The confidence level is 95% (CL = 0.95) CL = 0.95
so α = 1 – CL = 0.05
α/2 = 0.025 Z(α/2) = z0.025
The area to the right of Z0.025 is 0.025 and the area to the left of Z0.025 is 1 – 0.025 = 0.975
Z(α/2) = z0.025 = 1.645 This can be found using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645)*(9)/(85^0.5)=1.6058 x - EBM = 38 – 1.6058 = 36.3941 x + EBM = 38 + 1.6058 = 39.6058
The 95% confidence interval is (36.3941, 39.6058).
The answer is the letter D
The value of 40.2 is within the 95% confidence interval for the mean of the sample

2) To find the confidence interval 
the sample mean x = 76 σ = 20; n = 102; 
The confidence level is 95% (CL = 0.95) CL = 0.95
so α = 1 – CL = 0.05
α/2 = 0.025 Z(α/2) = z0.025
The area to the right of Z0.025 is 0.025 and the area to the left of Z0.025 is 1 – 0.025 = 0.975
Z(α/2) = z0.025 = 1.645 This can be found using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645)*(20)/(102^0.5)=3.2575 x - EBM = 76 – 3.2575 = 72.7424 x + EBM = 76 + 3.2575 = 79.2575 
The 95% confidence interval is (72.7424 ,79.2575).
The answer is the letter A and the letter D
The value of 71.8 and 79.8 are outside the 95% confidence interval for the mean of the sample

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2 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 $s = \frac{\sigma}{\sqrt{n}}$