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zloy xaker [14]
3 years ago
14

Find all solutions for a triangle with B=36 degrees, b=19, and c=36. Round to the nearest tenth.

Mathematics
1 answer:
lawyer [7]3 years ago
8 0

we know that

Applying the law of sines

b/sin B=c/sin C-------> solve for sin C

sin C=(c/b)*sin B-----> sin C=(36/19)*sin 36

sin C=1.1137-------> the value of the sines can not be greater than 1

The side lengths and angle given cannot be used to create a triangle

therefore


with the given values ​​there is no solution to form some triangle

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One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )
Gemiola [76]

QUESTION 1

The given inequality is

|m|-2>0

We group like terms to get,

|m|>2


This implies that,

-m>2 or m>2.

We simplify the inequality to get,

m or m>2.

We can write this interval notation to get,

(-\infty,-2)\cup (2,+\infty).


QUESTION 2

|x-4|-3\:>\:5.

We group like terms to get,


|x-4|\:>\:5+3.


|x-4|\:>\:8

We split the absolute value sign to get,

-(x-4)\:>\:8 or x-4\:>\:8


This implies that,


x-4\: or x-4\:>\:8


x\: or x\:>\:8+4


x\: or x\:>\:12


We can write this interval notation to get,

(-\infty,-4)\cup (12,+\infty).


QUESTION 3

The given inequality is

|6+9x|\leq 24


We split the absolute value sign to obtain,

-(6+9x)\leq 24 or (6+9x)\leq 24


This simplifies to

6+9x\ge -24 and 6+9x\leq 24


9x\ge -24-6 and 9x\leq 24-6


9x\ge -30 and 9x\leq 18


x\ge -\frac{10}{3} and x\leq 2

-\frac{10}{3}\leq x\leq2

We write this in interval form  to get,

[-\frac{10}{3},2]


QUESTION 4

The given inequality is

|1-5a|>29

We split the absolute value sign to get,

-(1-5a)>29 or 1-5a>29

This simplifies to,

1-5a\: or 1-5a\:>\:29


This implies that,

-5a\: or -5a\:>\:29-1


-5a\: or -5a\:>\:28


a\:>\:6 or a\:

We write this in interval notation to get,

(-\infty,-\frac{28}{5})\cup (6,+\infty)















7 0
3 years ago
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