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koban [17]
1 year ago
7

A local middle school adopted a policy for school uniforms. Students can wear black pants or tan pants. They can wear a yellow s

hirt, a red shirt, a green shirt, or a white shirt. The tree diagram shows the possible outfit choices.


How many different choices does a student have when choosing a pair of pants and a shirt?

2

4

8

10
Mathematics
1 answer:
ValentinkaMS [17]1 year ago
5 0
The answer is 8 because you have black and tan pants so 2 pairs and then a green, red, white, and yellow shirt sooooo you have 4 choices of shirts to pair with your pants. So 2•4=8
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If cos(t) = 2/7 and t is in the 4th quadrant, find sin(t).
Yuliya22 [10]
We can use the Pythagorean Trigonometric Identity which says:
sin^2(t)+cos^2(t)=1

Since we need to find sin(t), we have to solve for it:
sin(t)= \sqrt{1-cos^2(t)}

Let's plug in the given cos(t) value:
sin(t) = \sqrt{1-cos^2( \frac{2}{7})}

And solve sin(t):
sin(t) = \sqrt{1- \frac{4}{49} } = \frac{x}{y} \sqrt{ \frac{49}{49}- \frac{4}{49} }

Simplify further:
sin(t) = \sqrt{ \frac{45}{49} } = \frac{ \sqrt{45} }{7} = \frac{ \sqrt{9*5} }{7}

And it all simplifies down to:
sin(t) = \frac{3 \sqrt{5} }{7}

Since it's in the 4th quadrant, the sin(t) value is going to be negative. So, your final answer is: 
sin(t) = - \frac{ 3\sqrt{5} }{7}

Hope this helps!
7 0
2 years ago
Which oh the following are among the five basic postulates of Euclidean geometry
igomit [66]

1. A straight line segment can be drawn joining any two points.

2. Any straight line segment can be extended indefinitely in a straight line.

3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.

4. All right angles are congruent.

5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.

8 0
1 year ago
The data set represents the number of rings each person in a room is wearing.
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If the data set represents the number of rings each person is wearing, being: 0,2,4,0,2,3,2,8,6, the interquartile range of the data is 2. Being, 4 as the Q1, 3 as the Q2 or median, and 6 as the Q3. Where the formula of getting the interquartile range is IQR= Q1-Q2.
3 0
2 years ago
Read 2 more answers
What number would you divide by to calculate the mean of 3, 4, 5, and 6?
Mademuasel [1]

Answer:

4

Step-by-step explanation:

Mean = sum of scores /number of scores

We have four scores given ,so, the sum of the scores given will be divided by the number of scores which is 4

4 0
2 years ago
Read 2 more answers
g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                      = [ 121.60-1.645 \times {\frac{6.36}{\sqrt{36} } } , 121.60+1.645 \times {\frac{6.36}{\sqrt{36} } } ]

                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0
2 years ago
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