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belka [17]
3 years ago
8

Please help! ! ! ! !

Mathematics
1 answer:
Vladimir [108]3 years ago
3 0
B is the answer how you do good
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Find the value of x in the isosceles triangle shown below.
Solnce55 [7]

Answer: B. x=10

Step-by-step explanation:

<u>Concept</u>

- If it is an isosceles triangle, then the altitude would also be the median of the base, meaning that it divides the base evenly.

- We are going to use the Pythagorean theorem which is a²+b²=c², where [a] and [b] are legs, and [c] is the hypotenuse.

<u>Given</u>

Hypotenuse (relative) = √74

Altitude (one leg) = 7

The other leg (half of the base) = x/2

<u>Solve</u>

a² + b² = c²

(7)² + (x/2)² = (√74)²

49 + x²/4 = 74

x²/4 = 25

x² = 100

x = 10

Hope this helps!! :)

Please let me know if you have any questions

8 0
3 years ago
If 75% is 60, what is the full cost?
galina1969 [7]
If 75% is 60 then:
0.75x=60
Solve for x by dividing both sides by 0.75:
\frac{0.75x}{0.75}=\frac{60}{0.75}
x=80
4 0
3 years ago
Read 2 more answers
A function of random variables used to estimate a parameter of a distribution is a/an _____.
stiks02 [169]

A function of random variables utilized to calculate a parameter of distribution exists as an unbiased estimator.

<h3>What are the parameters of a random variable?</h3>

A function of random variables utilized to calculate a parameter of distribution exists as an unbiased estimator.

An unbiased estimator exists in which the difference between the estimator and the population parameter grows smaller as the sample size grows larger. This simply indicates that an unbiased estimator catches the true population value of the parameter on average, this exists because the mean of its sampling distribution exists the truth.

Also, we comprehend that the bias of an estimator (b) that estimates a parameter (p) exists given by; E(b) - p

Therefore, an unbiased estimator exists as an estimator that contains an expected value that exists equivalent to the parameter i.e the value of its bias exists equivalent to zero.

Generally, in statistical analysis, the sample mean exists as an unbiased estimator of the population mean while the sample variance exists as an unbiased estimator of the population variance.

Therefore, the correct answer is an unbiased estimator.

To learn more about unbiased estimators refer to:

brainly.com/question/22777338

#SPJ4

4 0
1 year ago
A sum of money is placed at simple interest for 3 years at 10% annum and then the amount is invested for 3 years at the same rat
fgiga [73]

Answer:

Initial principal -$300,000

Investment amount after yr 3-$390,000

Final investment amount, after 6 yrs-$519,090

Step-by-step explanation:

Let X be the amount initially invested.

#The amount after 3 yrs of simple interest is calculated as:

I=PRT\\\\A=P+I\\\\A=X+X\times 0.1\times 3\\\\A=1.3X

#The amount after 5 yrs is calculated by compounding the amount after 3 yrs for 2 yrs at 10%:

A=P(1+i)^2\\\\471900=1.3X(1.1)^2\\\\471900=1.573X\\\\X=300000

Hence, the amount initially invested as $300,000

#Amount of invested after 6 yrs is therefore:

A=P(1+i)^n\\\\=471900\times1.1\\\\=519090

Hence, the total amount of the investment after 6 yrs is $519,090

#Substitute X in simple interest equation to find amount after 3 yrs of simple interest:

I=PRT\\\\A=P+I\\\\A=X+X\times 0.1\times 3\\\\A=1.3X\\\\=1.3\times 300000\\\\=390000

Hence, the total amount of the investment after first 3 yrs is $390,000

3 0
4 years ago
HELP!!!
ella [17]

Answer:

120

Step-by-step explanation:

We are given that the function for the number of students enrolled in a new course is f(x) = 4^{x}-1.

It is asked to find the average increase in the number of students enrolled per hour between 2 to 4 hours.

We know that the average rate of change is given by,

A = \frac{f(x)-f(a)}{x-a},

where f(x)-f(a) is the change in the function as the input value (x-a) changes.

Now, the number of students enrolled at 4 = f(4) = f(x) = 4^{4}-1 = 255 and the number of students enrolled at 2 = f(2) = f(x) = 4^{2}-1 = 15

So, the average increase A=\frac{f(4)-f(2)}{4-2} = A=\frac{255-15}{4-2} = A=\frac{240}{2} = 120.

Hence, the average increase in the number of students enrolled is 120.

3 0
4 years ago
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