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3 years ago

Seven values on the number line above are marked with letters. Match the opposites.

Mathematics

1 answer:

Natali [406]3 years ago

5 0

Answer:

Step-by-step explanation:

hope this helps yout just have to put the negative letters with the positive!!!

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2 years ago

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2 years ago

Hi can someone help me with the last two sets of shapes? Thank you so much! -Jenn

SOVA2 [1]

You are correct there's at least one obtuse angle

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3 years ago

A paint manufacturer uses a machine to fill gallon cans with paint (1 galequals128 ounces). The manufacturer wants to estimate

kvasek [131]

Answer:

a) To determine the minimum sample size we need to use the formula shown in the picture 1.

E is the margin of error, which is the distance from the limits to the middle (the mean) of the confidence interval. This means that we have to divide the range of the interval by 2 to find this distance.

E = 0.5/2 = 0.25

Now we apply the formula

n = (1.645*0.80/0.25)^2 = 27.7 = 28

The minimum sample size would be 28.

b) To answer the question we are going to make a 90% confidence interval. The formula is:

(μ - E, μ + E)

μ is the mean which is 127. The formula for E is shown in the picture.

E = 0.80*1.645/√8 = 0.47

(126.5, 127.5)

This means that the true mean is going to be contained in this interval 90% of the time. This is why it doesn't seem possible that the population mean is exactly 128.

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3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

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3 years ago