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3 years ago

If g(x) = 4 square root of x then g(45) is

1 answer:

5 0

Hi,

So we have g(x) = 4√(x). We're looking for g(45). Think of it like this: whatever number is in the place of x in g(x), just place that number AS x.

Therefore, we have :
g(45) = 4√(45) ⇒ (4) × (√(45)) ⇒ (4)(3√(5)) = 12√(5).

-Hope this helps!

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Calculate the specified probability Suppose that K is a random variable. Given that P(-2.2 ≤ K ≤ 2.2) = 0.725, and that P(K <

belka [17]

Answer:

P(K > 2.2) = 0.1375

Step-by-step explanation:

We have that:

P(-2.2 ≤ K ≤ 2.2) = 0.725

P(K < -2.2) = P(K > 2.2)

It means that the distribution is symmetric.

The sum of all the probabilities is decimal 1.

We have the following probabilities:

P(K < 2.2)

P(-2.2 ≤ K ≤ 2.2)

P(K > 2.2)

P(K < 2.2) + P(-2.2 ≤ K ≤ 2.2) + P(K > 2.2) = 1

Since P(K < 2.2) = P(K > 2.2)

P(K > 2.2) + P(-2.2 ≤ K ≤ 2.2) + P(K > 2.2) = 1

2P(K > 2.2) + 0.725 = 1.

2P(K > 2.2) = 1 - 0.725

2P(K > 2.2) = 0.275

P(K > 2.2) = 0.275/2

P(K > 2.2) = 0.1375

6 0

3 years ago

Y=2+ square root of x-8 solve for x

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

y = 2√(x-8)

y² = 4(x-8) = 4x-32

4x = y²+32

x = 0.25y² +8

5 0

3 years ago

Help plz brainliest for first answer and best answer

Leto [7]

Answer:

C.15

Step-by-step explanation:

one box has 5 chocolates so 3*5=15

7 0

3 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$