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3 years ago

64-144a+108a^2-27a^3

Mathematics

1 answer:

Nataly [62]3 years ago

5 0

Answer:

= -27a^3+108a^2-144a+64

Step-by-step explanation:

Simplify:

=64-144a+108a^2-27a^3

=64+-144a+108a^2+-27a^3

Answer

= -27a^3+108a^2-144a+64

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Find the surface area of a cylinder with a base radius of 5 ft and a height of 6 ft.

Nadusha1986 [10]

Answer: 110pi

Solution: S = 2•pi•r•h + 2 pi•r^2 =2•6•5pi + 2•pi•25 = 60pi + 50pi = 110pi

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2 years ago

How to find the maximum height projectile motion?

ziro4ka [17]

Y=v(initial)*t+1/2at^2
since the velocity is zero at the maximum height then we get
y=1/2at^2
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so y=1/2gt^2.

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3 years ago

You are flying a kite on a line that is 350 feet long. Let's suppose the line is perfectly straight (it never really is) and it

aliina [53]

Answer:

We can think that the line of the kite is the hypotenuse of a triangle rectangle, and the altitude is one of the cathetus of the triangle.

And we know that it makes an angle of 65° with the horizontal (i guess this is measured between the hypotenuse and the horizontal adjacent to the kite.

This angle is complementary to the top angle of our triangle rectangle, such that A + 65° = 90°

A = 90° - 65° = 25°

Then the altitude of the kite is the adjacent cathetus to this angle.

We can use the relation:

sin(A) = Adjacent cathetus/hypotenuse.

Sin(25°) = X/350ft

Sin(25°)*350ft = X = 147.9m

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3 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

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3 years ago

» What is the combined length of all the buttons longer than 1 inch

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Answer:

how many buttons there are

Step-by-step explanation:

yo have to have a certain number of buttons

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2 years ago

64-144a+108a^2-27a^3​

64-144a+108a^2-27a^3