Please help me with number 8a and b

Tarin organized this table to determine the number of adults a and children attending a fundraiser based on the total number of

GarryVolchara [31]

Answer:

I don't see any answer choices.

6 0

2 years ago

Read 2 more answers

Find the mean of the following data set. 42, 45, 58, 63 42 47 52

igomit [66]

52 i believe.

42 + 45 + 58 + 63 = 208
208 / 4 = 52

6 0

2 years ago

Lin solved the equation 8(x-3)+7=2x(4-17) incorrectly. Find the errors in her solution. What should her answer have been?

Irina-Kira [14]

Answer: The correr solution is $x=\frac{17}{34}$

Step-by-step explanation:

Since the steps in Lin's solution are not shown, we are not able to find her erros. However, we can solve it step by step and find the correct result:

$8(x-3)+7=2x(4-17)$

Firstly, in the left part of the equation we have to multiply $8$ by the numbers inside the parenthesis; and in the right side we have to solve first what is inside the parenthesis:

$8x-24+7=2x(-13)$

Adding similar terms in the left side and multiplying $2x$ by the number inside the parenthesis in the right side:

$8x-17=-26 x$

Isolating $x$ :

$x=\frac{17}{34}$ This is the correct solution

4 0

3 years ago

An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an

igomit [66]

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

$P(e_1)=0.90*0.67=0.603$

2) Left knee unsuccessful and right knee successful (X=1).

$P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067$

3) Right knee unsuccessful and left knee successful (X=1).

$P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297$

4) Both surgeries are unsuccessful (X=0).

$P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033$

Then, the mean can be calculated as the expected value:

$M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57$

The variance can be calculated as:

$V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111$

3 0

3 years ago

When s is the open hemisphere x 2 + y 2 + z 2 = 1, z ≤ 0 , oriented by the inward normal pointing to the origin, then the bounda

Anit [1.1K]

The figure below shows a diagram of this problem. First of all we graph the hemisphere. This one has a radius equal to 1. Given that z ≤ 0 a sphere will be valid only in the negative z-axis, that is, we will get a half of a sphere that is the hemisphere shown in the figure. We know that this hemisphere is oriented by the inward normal pointing to the origin, then we have a Differential Surface Vector called N, using the Right-hand rule the boundary orientation is counterclockwise.

Therefore, the answer above False

5 0

3 years ago