Number of employees outside of company=40-29=11
we can calculate the percent decrease in staff by the rule of 3
40 employees-----------------100%
11 employees---------------- x
x=(11 employees*100%) / 40 employees=27.5%
The percent decrease in staff was: 27.5%
Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.
We do not know what x and y stand for. Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2 (common difference is positive)
Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}
Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To find the probability of all sockets in the sample being defective, we can do the following:
The first socket will be in a group where 5 of the 38 sockets are defective, so the probability is 5/38
The second socket will be in a group where 4 of the 37 sockets are defective, as the first one picked is already defective, so the probability is 4/37
Expanding this, we have that the probability of having all 5 sockets defective is: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%
b) Following the same logic of (a), the first socket have a chance of 33/38 of not being defective, as we will pick it from a group where 33 of the 38 sockets are not defective. The second socket will have a chance of 32/37, and so on.
The probability will be (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%
Answer:
20%
Step-by-step explanation:2.8-3.5 then divide that by 3.5 and multiple the answer by 100 to get the percent
Answer:
The variance in weight is statistically the same among Javier's and Linda's rats
The null hypothesis will be accepted because the P-value (0.53 ) > ∝ ( level of significance )
Step-by-step explanation:
considering the null hypothesis that there is no difference between the weights of the rats, we will test the weight gain of the rats at 10% significance level with the use of Ti-83 calculator
The results from the One- way ANOVA ( Numerator )
with the use of Ti-83 calculator
F = .66853
p = .53054
Factor
df = 2 ( degree of freedom )
SS = 23.212
MS = 11.606
Results from One-way Anova ( denominator )
Ms = 11.606
Error
df = 12 ( degree of freedom )
SS = 208.324
MS = 17.3603
Sxp = 4.16657
where : test statistic = 0.6685
p-value = 0.53
level of significance ( ∝ ) = 0.10
The null hypothesis will be accepted because the P-value (0.53 ) > ∝
where Null hypothesis H0 = ∪1 = ∪2 = ∪3
hence The variance in weight is statistically the same among Javier's and Linda's rats
