Hey can you please help me out posted picture

Answer: 12√3 square inches

Step-by-step explanation:

By the property of equilateral triangle,

Apothem = √3/2 × Side

⇒ Side = 2/√3 × Apothem

Here, apothem = 6 inches

Thus, the side of the given equilateral triangle =  [tex]\frac{2}{\sqrt{3}}\times 6[/tex]

= [tex] \frac{12}{\sqrt{3}}[/tex]

= [tex]4\sqrt{3}[/tex]  unit

Since, For an equilateral triangle,

[tex]\text{ Area} = \frac{\sqrt{3}}{4}\times (\text{ side})^2[/tex]

⇒ The area of the given equilateral triangle = [tex] \frac{\sqrt{3}}{4}\times (4\sqrt{3})^2[/tex]

[tex]=\frac{\sqrt{3}}{4}\times 48[/tex]

[tex]=12\sqrt{3}[/tex]  square inches

Answer:

There can be only one solution

Step-by-step explanation:

The only soution is zero, because anything else is unequal.

Answer:

y = 2 (x - 8 )2 + (-72)) .

The x-coordinate of the minimum is 8.

Step-by-step explanation:

I just took this test on plato and I got it correct.

23 photos per row. Got it? Good.

The result of 3 times as much as the sum of 3/4 and 2/6 is; 13/4

First, we must evaluate the sum of 3/4 and 2/6; we have;

Using the lowest common multiple; 12

We have; (9 +4)/12 = 13/12.

Therefore, 3 times 13/12 = 39/12 = 13/4

Read more on fraction addition;

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Answer:

B

Step-by-step explanation:

Answer:

Standard deviation of the distribution of sample means = 7.0855

Step-by-step explanation:

We are given that a sample of size 44 is drawn from a population with mean 36 and standard deviation 47.

Using Central Limit Theorem, it is stated that;

Standard deviation of the distribution of sample means = [tex]\frac{Population S.D.}{\sqrt{n} }[/tex]

         =  [tex]\frac{\sigma}{\sqrt{n} }[/tex]  = [tex]\frac{47}{\sqrt{44} }[/tex] = 7.0855

Answer:

Standard error = 5.4

Step-by-step explanation:

We are given that Population Mean, [tex]\mu[/tex] = 150 and Population Standard deviation, [tex]\sigma[/tex] = 27  and sample size, n = 25

Standard error formula for the sampling distribution of this sample mean is given by;

 Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{27}{\sqrt{25} }[/tex] = 5.4

Answer:

x=-4,1,2+5i,2-5i

Step-by-step explanation:

Given is an algebraic expression g(x) as product of two functions.

Hence solutions will be the combined solutions of two quadratic products

[tex]g(x) = (x^2 + 3x - 4)(x^2 - 4x + 29)\\[/tex]

I expression can be factorised as

[tex](x+4)(x-1)[/tex]

Hence one set of solutions are

x=-4,1

Next quadratic we cannot factorize

and hence use formulae

[tex]x=\frac{4+/-\sqrt{16-116} }{2} =2+5i, 2-5i[/tex]

Answer:

The answer above is correct:

B. 1

C. 4

E. 2+5i

F. 2-5i

Step-by-step explanation:

I got this right on Edg. 2021

Polygons that encircle a circle contain tangents of similar length, as such the perimeter of the polygon that circumscribes the circle is 76 cm.

The act of circumscribing a circle inside a polygon involves the process of drawing a circle that perfectly fits in the polygon by bisecting two arcs on each side of the polygon and then drawing a circle where all the sides meet in the middle of the polygon.

Polygons that encircle a circle contain tangents of similar length that originate at the same vertex, which explains the computation of the dimensions.

Taking a look at the figure attached, we have:

Thus, the perimeter of the polygon can be calculated as:

= (19 + 19 + 6 + 6 + 4 + 4 + 9 + 9) cm

= 76 cm

Learn  more about the perimeter of a polygon here:

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Approximately 0.2266, or 22.66%, is the probability that rolling the die more than 140 times is needed to exceed a total of 400.

To approximate the probability that rolling the die more than 140 times is needed to exceed a total of 400, we can use a normal approximation to the binomial distribution since the number of rolls is large.

First, let's calculate the mean (μ) and standard deviation (σ) of the number of rolls needed to exceed 400:

[tex]\[ \text{Mean (μ)} = \frac{\text{Total target}}{\text{Expected value per roll}} = \frac{400}{\frac{7}{2}} \][/tex]

[tex]\[ \text{Standard deviation (σ)} = \sqrt{\frac{\text{Total target} \times (\text{Sides}^2 - 1)}{12}} = \sqrt{\frac{400 \times (6^2 - 1)}{12}} \][/tex]

Now, we'll use the normal approximation and the z-score formula to find the probability:

[tex]\[ z = \frac{\text{X} - \text{μ}}{\text{σ}} \][/tex]

[tex]\[ z = \frac{140 - \text{μ}}{\text{σ}} \][/tex]

Then, we look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with that z-score.

Let's calculate these values.

First, let's calculate the mean (μ) and standard deviation (σ):

[tex]\[ \text{Mean (μ)} = \frac{400}{\frac{7}{2}} = \frac{800}{7} \approx 114.29 \][/tex]

[tex]\[ \text{Standard deviation (σ)} = \sqrt{\frac{400 \times (6^2 - 1)}{12}} = \sqrt{\frac{400 \times 35}{12}} \approx \sqrt{\frac{14000}{12}} \approx \sqrt{1166.67} \approx 34.16 \][/tex]

Now, let's find the z-score for rolling the die more than 140 times:

[tex]\[ z = \frac{140 - \text{μ}}{\text{σ}} = \frac{140 - 114.29}{34.16} \approx \frac{25.71}{34.16} \approx 0.75 \][/tex]

Using a standard normal distribution table or calculator, we find the probability associated with a z-score of 0.75, which represents the probability that rolling the die more than 140 times is needed to exceed a total of 400.

The Correct Question is :

A six-sided die, in which each side is equally likely to appear, is repeatedly rolled until the total of all rolls exceeds 400. Approximate the probability that this will require more than 140 rolls.

The approximate probability that it will require more than 140 rolls for the total to exceed 400 is 0.005.

To find the approximate probability that it will require more than 140 rolls for the total to exceed 400, we can relate it to the probability that the sum of the first 140 rolls is less than 400.

Let X be the random variable representing the sum of the rolls. We want to find P(X > 400), which is the probability that it will require more than 140 rolls.

We can calculate this by finding the complement of the event that the sum of the first 140 rolls is less than 400.

Let A be the event that the sum of the first 140 rolls is less than 400. Then, P(A) is the probability that we're interested in.

Now, we calculate P(A):

Since each side of the die is equally likely, the expected value of the roll is [tex]\( \frac{1+6}{2} = 3.5 \)[/tex].

The expected value of the sum of the first 140 rolls is [tex]\( 140 \times 3.5 = 490 \)[/tex].

Therefore, P(A) can be approximated using the normal distribution, since the sum of the rolls follows approximately a normal distribution due to the Central Limit Theorem.

Using the properties of the normal distribution, we can standardize the value:

[tex]\[ Z = \frac{400 - 490}{\sqrt{140 \times \left(\frac{1}{12}\right)}} \][/tex]

Here, [tex]\( \frac{1}{12} \)[/tex] is the variance of a single roll of the die.

Now, we find P(A) using the standardized value of Z:

P(A) = P(X < 400) = P(Z > z)

We can then find the probability from a standard normal distribution table or calculator.

[tex]\[ P(A) \approx P(Z > -2.589) \][/tex]

From a standard normal distribution table, we find that [tex]\( P(Z > -2.589) \approx 0.995 \)[/tex].

So, the approximate probability that it will require more than 140 rolls for the total to exceed 400 is 1 - 0.995 = 0.005.

The probable question may be:

A six-sided die, in which each side is equally likely to appear, is repeatedly rolled until the total of all rolls exceeds 400. What is the approximate probability that this will require more than 140 rolls? (Hint: Relate this to the probability that the sum of the first 140 rolls is less than 400.)