Hi can you help me It’s a quizz

A right cylinder has a height of 20 1/2ft and a diameter of 1 1/5 times it's height what is the volume of the cylinder? Use 3.14

anygoal [31]

Volume (V) of a cylinder is the area of circular base (pi×r^2) times it's length/height (h):
$V = \pi {r}^{2} (h) \: \: \pi = 3.14$
h = 20.5 ft
diameter (d) = 1 1/5 × h, 1 1/5 = 6/5 = 1.20
d = 1.20 × 20.5 = 24.6 ft
radius (r) = 1/2 d = 24.6/2 = 12.3 ft
$V = \pi {r}^{2} (h) \\ V = 3.14 ({12.3}^{2}) (20.5) \\ V = 3.14 (151.29) (20.5) \\ V = 475.05 (20.5) = 9738.54 \: {ft}^{3}$

4 0

3 years ago

The Highway Safety Department wants to study the driving habits of individuals. A sample of 37 cars traveling on a particular st

kodGreya [7K]

Answer:

90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

Step-by-step explanation:

We are given that a sample of 37 cars traveling on a particular stretch of highway revealed an average speed of 70.7 miles per hour with a standard deviation of 6.3 miles per hour.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample average speed of cars = 70.7 miles per hour

s = sample standard deviation = 6.3 miles per hour

n = sample of cars = 37

$\mu$ = true mean speed

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the true mean, $\mu$ is ;

P(-1.688 < $t_3_6$ < 1.688) = 0.90 {As the critical value of t at 36 degree of

freedom are -1.688 & 1.688 with P = 5%}

P(-1.688 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.688) = 0.90

P( $-1.688 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.688 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.688 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $70.7-1.688 \times {\frac{6.3}{\sqrt{37} } }$ , $70.7+1.688 \times {\frac{6.3}{\sqrt{37} } }$ ]

= [68.9517 miles per hour , 72.4483 miles per hour]

Therefore, 90% confidence interval for the true mean speed of all cars on this particular stretch of highway is [68.9517 miles per hour , 72.4483 miles per hour].

The interpretation of the above interval is that we are 90% confident that the true mean speed of all cars will lie between 68.9517 miles per hour and 72.4483 miles per hour.

8 0

3 years ago

Solve for the 3 variables

Ipatiy [6.2K]

I realised I wasn't doing it right.

8 0

4 years ago

Hayden put 2 cups of lemon juice and 5 time as much water into a jar of lemonade. How many fluid ounces of lemonade were in the

olchik [2.2K]

Answer:

96 fluid ounces

Step-by-step explanation:

We have to convert cups to fluid ounces

1 cup = 8 fluid ounces

8 fluid ounces = 8 x 2 = 16 fluid ounces

the quantity of water = 16 x 5 = 80 fluid ounces

Total fluid ounces = 80 + 16 = 96 fluid ounces

4 0

3 years ago

You are given the information that P(A) = 0.30 and P(B) = 0.40.

Ad libitum [116K]

Answer:

1.B. No. You need to know the value of P(A and B). 2.C. Yes P(A and B) =0, so P(A or B) = P(A) + P(B).

Step-by-step explanation:

We can solve this question considering the following:

For two mutually exclusive events:

$\\ A_{1}\;and\;A_{2}$

$\\ P(A_{1} or A_{2}) = P(A_{1}) + P(A_{2})$ (1)

An extension of the former expression is:

$\\ P(A_{1} or A_{2}) = P(A_{1}) + P(A_{2}) - P(A_{1} and A_{2})$ (2)

In mutually exclusive events, P(A and B) = 0, that is, the events are independent one of the other, and we know the probability that both events happen at the same time is zero (P(A and B) = 0). There are some other cases in which if event A happens, event B too, so they are not mutually exclusive because P(A and B) is some number different from zero. Notice the difference between OR and AND. The latter implies that both events happen at the same time.

In other words, notice that the formula (2) provides an extension of formula (1) for those events that are not mutually exclusive, that is, there are some cases in which the events share the same probabilities in a way that these probabilities must be subtracted from the total, so those probabilities in common do not "inflate" the actual probability.

For instance, imagine a person going to a gas station and ask for checking both a tire and lube oil of his/her car. The probability for checking a tire is P(A)=0.16, for checking lube oil is P(B)=0.30, and for both P(A and B) = 0.07.

The number 0.07 represents the probability that both events occur at the same time, so the probability that this person ask for checking a tire or the lube oil of his/her car is:

P(A or B) = 0.16 + 0.30 - 0.07 = 0.39.

That is why we cannot simply add some given probabilities without acknowledging if the events are or not mutually exclusive, whereas we can certainly add the probabilities in question when we know that both probabilities are mutually exclusive since P(A and B) = 0.

In conclusion, knowing the events are mutually exclusive does provide extra information and we can proceed to simply add the probabilities of either event; thus, the answers are those in which we need to previously know the value of P(A and B).

7 0

3 years ago