What is the standard form of 65

I need help please?!):

motikmotik

Answer:

Option C: $\sqrt{\textbf{121}}$

Step-by-step explanation:

Given the side length of a square is 11 units.

Note that $\sqrt{121} = \pm 11$

Since, length cannot be negative, we can eliminate - 11.

Hence, the length of the side of the square is 11 units = $\sqrt{11^2} = \sqrt{121}$ .

Hence, the answer.

7 0

3 years ago

Substance X is found in the bloodstream of humans. A random sample of 75 males had a mean content of 28 ppm of Substance X in th

Mashcka [7]

Answer:

H0: u1 - u2 = 0

H1: u1 - u2 ≠ 0

Step-by-step explanation:

This question requires us to state the hypothesis to be used for this test to know if males and females have the same amount of substance X. I have gone further to also calculate the test statistics too.

Null hypothesis:

H0: u1 - u2 = 0

Alternative hypothesis:

H1: u1 - u2 ≠ 0

X1 = 28, sd1 = 14.1, n1 = 75

X2 = 33, sd2 = 9.5, n2 = 50

We have a 2 tailed test

Alpha = 1-0.8812

= 0.1188

Critical value = 1.56

If z<-1.56 or z>1.56

Reject null hypothesis

Test statistic

= (28-33)/14.1²/75 + 9.5²/50

= -2.37

Therefore we reject the H0 because -2.37 is less than -1.56

5 0

3 years ago

Nina is 10 years younger than Deepak. Deepak is 3 times as old an Nina. Which system of equations can be used to find d, Deepak’

andre [41]

The first choice is the correct answer.

8 0

4 years ago

Read 2 more answers

The inequality x<4 in interval notation

Strike441 [17]

X<4 represents "all numbers smaller than 4, including all negative numbers."

In interval notation, this comes out to (-infinity, 4).

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

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

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

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

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

3 0

4 years ago