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2 years ago

Mr. Hou put students names in a hat and selected file names without looking. What type of sample did he preform?

Mathematics

1 answer:

Anettt [7]2 years ago

8 0

Step-by-step explanation:

The answer is A. The answer is A because the teacher put names in a hat without looking. That narrows down the answer to A and B. Since he didn't do it in a special scientific way the answer is A

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What is an array of 4 x 875

OLEGan [10]

4 rows and 875 columns, or 875 rows and 4 columns

5 0

2 years ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

2 years ago

The system shown has the unique solution (2, y, z). Solve the system and select the values that complete the solution. y = 0 y =

madreJ [45]

So we are given a system:
$3x-2y+3z=0\\
-3x - 5y - 5z= -21$
Substitute x = 2 we get the system:
$-2y+3z=-6\\
- 5y - 5z= -15$
Multiply the first equation by -5 and the second by 2 we get the system:
$10y-15z=30\\
- 10y - 10z= -30$
Adding the two equations we get :
$-25z=0\text{ then}z=0.$
We find the value of y by using any of the other equations like this:
$-2y=-6\\y=3.$
Final solution:
$z=0,y=3$

8 0

2 years ago

Read 2 more answers

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago

An American car travels 32 miles on one gallon of gas. A European car travels 12.7 kilometers on one liter of gas. Which car get

BabaBlast [244]

32 miles is the same as 19.8839 kilometers

There are 3.78541 liters in a gallon

Now we divide the two.

19.8839 / 3.78541 ≈ 5.25

The American car gets 5.25 km per liter. So the European car is better on gas economy.

5 0

2 years ago