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nirvana33 [79]
2 years ago
13

Help me please i beg you

Mathematics
1 answer:
k0ka [10]2 years ago
8 0
ANSWER: Acute and obtuse angle
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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
2 years ago
Which expression is equivalent to 24^1/3
kenny6666 [7]
Second is right answer
3 0
3 years ago
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What is an example of a one step equation that would use multiplication
poizon [28]

Answer:

Multiplication equation 6 t = 54 6t = 54 6t=54 Multiply each side by five.

7 0
2 years ago
A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra
Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

z=1

\frac{dx}{dt}=430

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

y=2

According to Pythagoras

hypotenuse^2=base^2+perpendicular^2

y^2=x^2+1^2

y^2=x^2+1               .... (1)

Put z=1 and y=2, to find the value of x.

2^2=x^2+1^2

4=x^2+1

4-1=x^2

3=x^2

Taking square root both sides.

\sqrt{3}=x

Differentiate equation (1) with respect to t.

2y\frac{dy}{dt}=2x\frac{dx}{dt}+0

Divide both sides by 2.

y\frac{dy}{dt}=x\frac{dx}{dt}

Put x=\sqrt{3}, y=2, \frac{dx}{dt}=430 in the above equation.

2\frac{dy}{dt}=\sqrt{3}(430)

Divide both sides by 2.

\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}

\frac{dy}{dt}=372.390923627

\frac{dy}{dt}\approx 372

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0
3 years ago
Can someone help me in this question?
igomit [66]

Answer:

3347000 cm is 33.47 km to cm

3347000/64=

401.92 is the circumference

since it is pi * d

do 3347000/401.92

8327.52786624 round to

8327 full revolutions

3 0
2 years ago
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