Let's find the rates at which Melinda and Marcus are saving money and compare them.
Based on the table, we can see that Melinda originally had $75. Then she got $135 in 5 weeks. So the rate of saving money is (135–75)/5 = $12 per week.
This rate is unchanged for the next weeks. As we can see, she got $195 in the next 5 weeks. So she saved $60 more those 5 weeks, or the rate is $60/5 = $12 per week again.
So Melinda saved $12 per week.
As for Marcus, the equation tells us that the rate of saving money is $14 which is the coefficient in front of x.
Hence, t<span>he rate at which Melinda is adding to her savings each week is 2$
less than the rate at which Marcus is adding to his savings each week.</span>
R^2+(ab)^2= (ao)^2
ab=6
ao=11.7
Plug in
r^2+6^2=11.7^2
simplify
r^2+36= 136.89
-36 both sides
r^2=100.89
square root both sides
r= 10.04 rounded 10
Answer:
37 9/20
Step-by-step explanation:
Answer:
14 and 16.
Step-by-step explanation:
We know that two consecutive even numbers is 30.
Let's let n be <em>any number</em>.
Then the first <em>even</em> number must be 2n.
This is because n can be any number, either even or odd. However, if we multiply it by 2, this <em>ensures</em> that 2n is even. Remember that any number multiplied by 2 yields an even number.
Therefore, the consecutive even number will be (2n+2). Not (2n+1), because that gives an odd number.
We know that they sum to 30. So, we can write the following equation.

And we solve from there:

So, the value of n is 7.
Therefore, the first even number is 7(2)=14.
And the consecutive even number is 16.
Edit: Typo
Solution :
a).
Given : Number of times, n = 25
Sigma, σ = 0.200 kg
Weight, μ = 13 kg
Therefore the hypothesis should be tested are :


b). When the value of 
Test statics :



= 45.5
P-value = 2 x P(Z > 45.5)
= 2 x 1 -P (Z < 45.5) = 0
Reject the null hypothesis if P value < α = 0.01 level of significance.
So reject the null hypothesis.
Therefore, we conclude that the true mean measured weight differs from 13 kg.