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3 years ago

Is this linear or not linear? Please explain how you got your answer.

Mathematics

2 answers:

Viktor [21]3 years ago

8 0

Linear is the correct answer

Maru [420]3 years ago

6 0

Answer:

linear

Step-by-step explanation:

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I need help with these 2 questions!! please 20 points!!!

bezimeni [28]

9514 1404 393

Answer:

45 cm²
38 cm²

Step-by-step explanation:

The conventional way to work these problems is to make use of the formulas for areas of a triangle, rectangle, and semicircle. If you've used these formulas for a while, you recognize that they give you certain relationships that may make these problems easy to do mentally.

1. The area of a triangle is half the product of height and width, so is equivalent to the area of a rectangle either half as high, or half as wide. We note that the width of the sail is 5 cm, so half that is 2.5 cm--exactly the same as the height of the boat's hull. That means we can add the height of the sail to the length of the boat, and the total area can be considered to be the same as that of a rectangle that is

2.5 cm high × (12 cm + 6 cm) long = (2.5)(18) cm² = 45 cm²

2. The usual formula for the area of a circle is ...

A = πr²

When expressed in terms of diameter, this becomes ...

A = π(d/2)² = (π/4)d²

Then the area of a semicircle is half that, or (π/8)d². This is equivalent to the area of a rectangle that is "d" wide and "π/8·d" high. That is, the approximate area of the semicircle is that of a rectangle 4 cm high by (π/8·4 cm) = π/2 cm wide. In other words, the semicircle effectively adds π/2 cm to the left end of the 6 cm central rectangle of the figure.

As discussed above, the area of a triangle is equivalent to the area of a rectangle half as high. In this figure, the triangle is 10-6 = 4 cm wide, so can be considered to contribute 4/2 = 2 cm to the right end of the 6 cm central rectangle.

If we consider π ≈ 3, then the approximate area of the figure is ...

(4 cm)(3/2 cm + 6 cm + 2 cm) = (4)(9.5) cm² = 38 cm²

The exact value is 4(8+π/2) = 32+2π ≈ 38.283 cm².

7 0

2 years ago

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti

jeyben [28]

Answer:

We need a sample of size at least 13.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}$

$\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}$

$n = 12.21$

Rounding up

We need a sample of size at least 13.

3 0

2 years ago

Look at screenshot for more info

podryga [215]

C because I just did that

8 0

2 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

6 0

9 months ago

What’s the mean of 55,60,60,40,60,75,40,25

Mazyrski [523]

Answer:

51.625

Step-by-step explanation:

you add them all up to get 415 then you divide by eight because there are eight numbers that you add up to get the answer 51.625

8 0

3 years ago

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