Answer:
38% order a chicken dinner
Step-by-step explanation:
40% order tacos. 50% of 40% order chicken tacos. 50% of 40% = 0.5 * 0.4 = 0.2 = 20% order chicken tacos.
60% order fajitas. 30% of 60% order chicken fajitas. 30% of 60% = 0.3 * 0.6 = 0.18 = 18% order chicken fajitas.
20% + 18% = 38% order a chicken dinner.
Answer:
Step-by-step explanation:
Answer:
x = 10°
Step-by-step explanation:
a). Since, opposite angles of a cyclic quadrilateral are supplementary angles"
Therefore, in cyclic quadrilateral ABDE,
m∠ABD + m∠AED = 180°
110° + m∠AED = 180°
m∠AED = 180° - 110°
= 70°
b). AD = ED [Given]
m∠EAD = m∠AED [Since, opposite angles of equal sides are equal in measure]
m∠EAD = m∠AED = 70°
By triangle sum theorem in ΔABD,
m∠BAD + m∠ABD + m∠ADB = 180°
m∠BAD + 110° + 40° = 180°
m∠BAD = 180 - 150
= 30°
m∠AEB = m∠AED + m∠DAB [By angles addition postulate]
m∠AEB = 70° + 30°
= 100°
By triangle sum theorem in the large triangle,
x° + m∠AEB + m∠EAB = 180°
x° + 100° + 70° = 180°
x = 180 - 170
x = 10°
When dividing same numbers with different exponents (they’re both 10 so this works different than if they were different numbers) you subtract the top exponent by the bottom exponent. So 15-4=11 ultimately making the solution 10^11. Hope this helps :)
Answer:
d. t distribution with df = 80
Step-by-step explanation:
Assuming this problem:
Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:
a. t distribution with df = 82.
b. t distribution with df = 81.
c. t distribution with df = 41.
d. t distribution with df = 80
Solution to the problem
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with 
degrees of freedom and the pooled variance 
is given by this formula:
This last one is an unbiased estimator of the common variance 
So on this case the degrees of freedom are given by:
And the best answer is:
d. t distribution with df = 80
