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IgorC [24]
2 years ago
13

An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50

ft per sec. The height s of the ball in feet is given by the equation s =-27t2 + 50t+ 6.5, where t is the number of seconds after the ball was thrown
a) what is the initial height of the projectile?
b) what is the maximum height of the projectile? ( round to the nearest foot)
c) after how many second is the ball 12 feet above the moons surface? ( round to the nearest foot)
d) how long will it take for the ball to hit the moons surface?( round to the nearest hundredth)
Mathematics
1 answer:
ASHA 777 [7]2 years ago
5 0

Answer:

please answer me binary number multiply 11010 multiply 101

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36-9 divided [(0.24*5)+ (0.66*5)]
Ray Of Light [21]
[(0.24*5)+ (0.66*5)] == 4.50
0.24x5 = 1.20
0.66x5 = 3.30
+ ____
4.50

27.00÷4.5 = 0.166666667
4 0
3 years ago
Pls answer this question I give brainliest thank you!
matrenka [14]

Answer:

(0-10)

Step-by-step explanation:

If she plans to use 10 intervals and the highest number in the data set is just under 100, then it makes sense to split the histogram into ten intervals of 10.

6 0
2 years ago
What's 12+89+75+54+43+23=? <br> LOL!!!
Sladkaya [172]

Answer:

296

Step-by-step explanation:

5 0
3 years ago
Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico
IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

                                  P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean nicotine content = 27.3 milligrams

            s = sample standard deviation = 2.8 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>Part I</u> : So, 99% confidence interval for the population mean, \mu is ;

P(-3.355 < t_8 < 3.355) = 0.99  {As the critical value of t at 8 degree

                                      of freedom are -3.355 & 3.355 with P = 0.5%}  

P(-3.355 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 3.355) = 0.99

P( -3.355 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X-3.355 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

<u />

<u>99% confidence interval for</u> \mu = [ \bar X-3.355 \times {\frac{s}{\sqrt{n} } } , \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 27.3-3.355 \times {\frac{2.8}{\sqrt{9} } } , 27.3+3.355 \times {\frac{2.8}{\sqrt{9} } } ]

                                          = [27.3 \pm 3.131]

                                          = [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

<u>Part II</u> : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

                                  P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean nicotine content = 24.9 milligrams

            s = sample standard deviation = 2.6 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

So, 98% confidence interval for the population mean, \mu is ;

P(-2.896 < t_8 < 2.896) = 0.98  {As the critical value of t at 8 degree

                                       of freedom are -2.896 & 2.896 with P = 1%}  

P(-2.896 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.896) = 0.98

P( -2.896 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.896 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u />

<u>98% confidence interval for</u> \mu = [ \bar X-2.896 \times {\frac{s}{\sqrt{n} } } , \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 24.9-2.896 \times {\frac{2.6}{\sqrt{9} } } , 24.9+2.896 \times {\frac{2.6}{\sqrt{9} } } ]

                                          = [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0
2 years ago
Which table shows y as a function of x
ANTONII [103]
The answer should be C :) in order for there to be a function, first rule is the x number cannot repeat
3 0
2 years ago
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