Answer:
It is c and d. If this helps you are welcome
Step-by-step explanation:
Answer:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
Step-by-step explanation:
A null hypothesis is a sort of hypothesis used in statistics that intends that no statistical significance exists in a set of given observations.
It is a hypothesis of no difference.
It is typically the hypothesis a scientist or experimenter will attempt to refute or discard. It is denoted by H₀.
Whereas, the alternate hypothesis is the contradicting statement to the null hypothesis.
The alternate hypothesis describes direction of the hypothesis test, i.e. if the test is left tailed, right tailed or two tailed.
It is also known as the research hypothesis and is denoted by H
ₐ.
In this case we need to test whether the amount is paid after the grace period, on average, more than 4 times in 2018.
The hypothesis can be defined as follows:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
18 kg of 15% copper and 72 kg of 60% copper should be combined by the metalworker to create 90 kg of 51% copper alloy.
<u>Step-by-step explanation:</u>
Let x = kg of 15% copper alloy
Let y = kg of 60% copper alloy
Since we need to create 90 kg of alloy we know:
x + y = 90
51% of 90 kg = 45.9 kg of copper
So we're interested in creating 45.9 kg of copper
We need some amount of 15% copper and some amount of 60% copper to create 45.9 kg of copper:
0.15x + 0.60y = 45.9
but
x + y = 90
x= 90 - y
substituting that value in for x
0.15(90 - y) + 0.60y = 45.9
13.5 - 0.15y + 0.60y = 45.9
0.45y = 32.4
y = 72
Substituting this y value to solve for x gives:
x + y = 90
x= 90-72
x=18
Therefore, in order to create 90kg of 51% alloy, we'd need 18 kg of 15% copper and 72 kg of 60% copper.
Answer:
23/4
Step-by-step explanation:
This means that it is 3/4 of the distance in the x <u>and</u> y coordinates, so it is 3/4 of the way from 2 to 7, and since this is a difference of 7-2=5, then it is 2 + (5 * 3/4) or 2 + 15/4 or 23/4.
7/15 or o.46 that is the answer to this question
