Answer:
a] 38.5$
b] 118.71$
Step-by-step explanation:
A)35$ was increased by 10%
So:
10% of 35$
10/100 x 35 ($)
3.5$
So 35$ was increased by 3.5$ and the new price is 38.5$
B)105.99$ was increased by 12%
So:
12% of 105.99$
12/100 x 105.99$
=12.7188$
Let's round it to the nearest hundredths:
12.72$
So 105.99$ was increased by 12.72$ and the new price is 118.71$
HOPE IT HELPS :D
Tip: "of" means multiplication in Maths.
Answer:
There is a 90.32% probability that the cake was baked by Doug.
Step-by-step explanation:
We have these following probabilities:
A 70% probability that Doug bakes the cake.
A 30% probability that Jeremy bakes the cake.
A 40% probability that a cake baked by Doug gets a thumbs up.
A 10% that a cake baked by Jeremy gets a thumbs up.
One cake was selected at random on 10/01/2014 and got a "thumbs up".
1. Find the probability that the cake was baked by Doug.
The probability that a baked cake gets a thumbs up is:
Of those, 0.7*0.4 = 0.28 are baked by Doug.
So the probability is:
There is a 90.32% probability that the cake was baked by Doug.
The water would last a very little amount of water
what's the question your tryna figure out
Answer: Z is less than Zc ∴ 1.342 < 1.96
Therefore, Null hypothesis is not Rejected.
There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.
Step-by-step explanation:
Given that;
U = 75
X = 78
standard deviation α = 10
sample size n = 20
population is normally distributed
PROBLEM is to test
H₀ : U = 75
H₁ : U ≠ 75
TEST STATISTIC
since we know the standard deviation
Z = (X - U) / ( α /√n)
Z = ( 78 - 75 ) / ( 10 / √20)
Z = 1.3416 ≈ 1.342
Now suppose we need to test at ∝ = 0.05 level of significance,
Then Rejection region for the two tailed test is Zc = 1.96
∴ Reject H₀ if Z > Zc
and we know that Z is less than Zc ∴ 1.342 < 1.96
Therefore, Null hypothesis is not Rejected.
There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.