Answer/Step-by-step explanation:
Use equation y = mx+b
m is the slope and in the place for m is 15
b is the y intercept and because there is no y-int. it would be 0
Since the y-int is 0 then it is proportional.
This would be a positive slope.
Hope this helps :)
Answer:
improper question
Step-by-step explanation:
-_-
Answer:
Step-by-step explanation:
Hello!
The researchers want to compare the weight of the basset hounds in 1915 with the weight of the basset hound in 2015.
Using a sample of 36 male basset hounds taken in 1915 and a sample of 36 male basset hounds taken in 2015 a 90% CI for the difference of mean weights of the basset hounds (2015-1915) was constructed:
Group 1
X₁: Weight of a basset hound measured in 1915
n₁= 36
Group 2
X₂: Weight of a basset hound measured in 2015
n₂= 36
Difference X₂ - X₁
X[bar]₂ - X[bar]₁= -2.8cm
margin of error d= 1.3cm
90% CI: [-1.4; -1.5]cm
The calculated CI is two-tailed, you can use it to decide over a two-tailed hypothesis test for the same parameter of interest over the same level. If the hypothesis is:
H₀: μ₂ - μ₁=0
H₁: μ₂ - μ₁≠0
α: 0.10
The CI doesn't contain the zero, so you could reject the null hypothesis. With this, you can conclude that the difference between the average weight of the basset hounds in 2015 and the average weight of the basset hounds in 1915.
Judging by the fact that both limits of the confidence interval are negative, as well as the difference between point estimators, it seems that the weight of the basset hounds in 2015 is less than their weight in 1915. Of course, you are not valid to make that kind of conclusion without a proper hypothesis test.
Answer:
20 ft
Step-by-step explanation:
Answer:
The amount invested at 4.5% was 
The amount invested at 5% was 
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
Let
x -----> the amount invested at 4.5%
9,000-x -----> the amount invested at 5%
in this problem we have
substitute
so
therefore
The amount invested at 4.5% was
The amount invested at 5% was
