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1 year ago

EASY POINTS Classify the triangle below by its angles and sides.

Mathematics

1 answer:

fomenos1 year ago

6 0

Answer:

equilateral

Step-by-step explanation:

equilateral sides are equal and the angles aswell

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The u.s. dairy industry wants to estimate the mean yearly milk consumption. a sample of 16 people reveals the mean yearly consum

lidiya [134]

1-2. The best estimate for the population mean would be sample mean of 60 gallons. Since we know that the sample mean is the best point of estimate. Since sample size n=16 is less than 25, we use the t distribution. Assume population from normal distribution.

3. Given a=0.1, the t (0.05, df = n – 1 = 15)=1.75

4. xbar ± t*s/vn = 60 ± 1.75*20/4 = ( 51.25, 68.75)

5. Since the interval include 63, it is reasonable.

3 0

3 years ago

ASAP Please help. Thank You!!! :)

liraira [26]

#25.
Convert the fractional numbers into decimals for easier comparison.
3 2/3 is approximately 3.66
-4 2/5 is approximately -4.4
We now have: 3, 3.66, -4.2, -4.4
In order: -4 2/5, -4.2, 3, 3 2/3

#26.
All positive numbers are greater than negative numbers.
When comparing two negative numbers, the one with the larger absolute value, will be less.
So (G) is correct

#27.
25 = m + 6.3
25 - 6.3 = m + 6.3 - 6.3
m = 18.7

#28.
h = 3.2-1.6 = 1.6

#29.
x = 15.6 - 9.8 = 5.8

#30
p = 17 - 4.5 = 12.5

6 0

4 years ago

Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p

scZoUnD [109]

Answer:

The probability that they will teach different courses is $\frac{2}{3}$ .

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here x represents the course taught by the first part-time instructor and y represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Compute the probability that they will teach different courses as follows:

$P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}$

Thus, the probability that they will teach different courses is $\frac{2}{3}$ .

3 0

3 years ago

Please answer this correctly

Ulleksa [173]

Answer:

37.5%

Step-by-step explanation:

So the original fraction is 3/8 and we can’t simplify it so we gotta turn it into a percent, and we can do that by dividing 3 and 8nto get .375 so the percent is 37.5%.

7 0

3 years ago

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A box contains 8 green, 4 yellow, and 8 purple balls.

Kruka [31]

Answer:

4/20=1/5

Step-by-step explanation:

5 0

3 years ago

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