Suppose the students who scored 85 and 90 on the math test take the test again and score 95. How many stars would you have to add to the picturegraph next to 95?
10

5 0

3 years ago

The linear functions f(x) and g(x) are represented on the graph, where g(x) is a transformation of f(x):

Brilliant_brown [7]

Answer:

Part A: Determine the test average for your math class after completing test 2. (2 points)

Part B: Determine the test average for your science class after the completing test 2. (2 points)

Part C: Which class had a higher average after completing test 4? Show work to support your answer. (6 points)

I am finding itz answer, pls help

7 0

2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

Two cargo ships spot a signal fire on a small island. The captains know they are 140 feet

sweet [91]

Answer:

add all the angles

Step-by-step explanation:

140+82+78

6 0

3 years ago