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2 years ago

Will Identify the outlier in the data set. 77, 73, 78, 78, 71, 75, 120, 75, 77, 72, 75, 70 The outlier is

Mathematics

1 answer:

o-na [289]2 years ago

7 0

Answer:

120

Step-by-step explanation:

120

every other number is in the 70s

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If a line has a slope of 2 and contains the point (-2,1), what is its equation in point-slope form?

kondaur [170]

The equation for slope is y=mx+b. slope is m. 2 is the slope. so you know the m part of the equation is 2. just by eliminating the answers that don’t have 2 as the slope, A is your answer.

5 0

2 years ago

When a person is breathing normally the amount of air in their lawns varies sinusoidally. When full Karen’s lungs hold 2.8 L of

makkiz [27]

Answer:

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

Step-by-step explanation:

Let the function of quantity in the lung of air be A(t)

So $A(t) \alpha \sin (\frac{t - \alpha }{k} )$

so, A(t) = Amax sin t + b

A(t) = 2.8t⇒ max

A(t) = 0.6t ⇒ min

max value of A(t) occur when sin(t) = 1

and min value of A(t) = 0

So b = 0.6

and A(max) = 2.2

$A(t) = 2.2\sin \frac{(t)}{k} + 0.6$

at t = 2 sec volume of a is 0.6

So function reduce to

$A(t) = 2.2\sin \frac{(t - 2)}{k} + 0.6$

and t = 5 max value of volume is represent

so,

$\sin \frac{t - \alpha }{k} = 1$

$\frac{t - 2}{k} = \frac{\pi }{2}$ when t = 5

$\frac{6}{\pi } = k$

so the equation becomes

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

7 0

2 years ago

What is the value of x? Enter your answer in the box. mm Triangle V T K with segment T Y such that Y is on segment V K, between

lapo4ka [179]

Answer:

VK = x = 98 mm

Step-by-step explanation:

TY is the bisector of angle T, so it divides segments of the triangle proportionally. That is, ...

VY/VT = KY/KT

VY = VT·KY/KT = (57 mm)·68/192.2 = 30 mm

Then

VK = VY + YK = 30 mm + 68 mm

VK = 98 mm

5 0

2 years ago

Read 2 more answers

Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int

meriva

Answer:

b) 95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are $(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })$ the 95 percent confidence interval for the population mean 'μ'