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3 years ago

Two friends share 1/3 of a pitcher of

Mathematics

1 answer:

Oxana [17]3 years ago

7 0

Answer:

⅙ pitcher

Step-by-step explanation:

½ of ⅓ pitcher = ½ × ⅓ pitcher = ⅙ pitcher

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Can someone help me please

Anon25 [30]

Answer:

$y = - \frac{2}{3} x + 7$

You get this equation when you plug in the values to find out the y intercept.

3 0

3 years ago

14. During math class, Hallie created the following

kykrilka [37]

Answer:

Step-by-step explanation:

work backwards so do the opposite of what they told you

example: - you do +

multiply, you do divide etc etc

7 0

2 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago

The quadrilateral shown is a rectangle. What is the measure of PR? (round to the nearest hundredth) A) 34.06 B) 37.12 C) 40.24 D

Nata [24]

Answer: OPTION B.

Step-by-step explanation:

The missing figure is attached.

For this exercise you need to use the Pythagorean Theorem. This is:

$a^2=b^2+c^2$

Where "a" is the hypotenuse and "b" and "c" are the legs of the triangle.

In this case you can identify in the figure the triangle PRS, so the diagonal PR is the hypotenuse of the triangle PRS.

Then you can say that:

$a=PR\\b=PS=17\\c=RS=33$

Therefore, knowing these values, you can substitute them into $a^2=b^2+c^2$ :

$PR^2=17^2+33^2$

Finally you must solve for PR in order to find its value. This is:

$PR=\sqrt{17^2+33^2}\\\\PR=37.12$

3 0

4 years ago

Is the product of a fraction that is greater than one and any whole number less than or greater than the whole number? Explain y

uysha [10]

F the fraction is greater than 1, in other words the numerator is greater than the denominator, the product will be greater than the whole number.If the fraction is less than one, in other words the numerator is less than the denominator, the product will be less than the whole number.If the fraction is equal to 1, in other words the numerator is equal to the denominator, the product will equal the whole number.

7 0

4 years ago