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3 years ago

A sprinkler waters in a circular pattern, spraying water out 3.2 meters. How much surface will it water

Mathematics

1 answer:

Zepler [3.9K]3 years ago

8 0

Answer:

The Area of a circle is

πr2=808.64

(3.14)r2=808.64

Divide both sides by 3.14

r2=257.5286624

Square root both sides to get radius

r=16 feet

Step-by-step explanation:

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Find quotient and remainder b)diving 2x^3- 5x² +x+2

tekilochka [14]

Given:

Dividend = $2x^3-5x^2+x+2$

Divisor = $x-2$

To find:

The quotient and remainder.

Solution:

We have,

Dividend = $2x^3-5x^2+x+2$

Divisor = $x-2$

Divide $2x^3-5x^2+x+2$ by $x-2$ as shown in the below picture.

Dividing $2x^3-5x^2+x+2$ by $x-2$ , we get

$Quotient=2x^2-x-1$

$Remainder=0$

Therefore, $Quotient=2x^2-x-1$ and $Remainder=0$ .

3 0

3 years ago

H = 3 + 31t - 161² Find all values of t for which the ball's height is 17 feet .

jasenka [17]

Answer:

t= 31/175 ≈ 5.64516129

Step-by-step explanation:

Refer Pic For Steps....

Hope it Helps!!!

3 0

2 years ago

How do you do these? 2x + 3> -9 & -3x + 5< 14

Alecsey [184]

First get everything to one side by adding 9 to 2x + 3 and you end up with 2x + 3 + 9 > 0 and you add 9 and 3 and it will be 2x + 12 > 0 and do the same for the second one but if your trying to solve for x, subtract the 3 on both sides and you will get 2x > -12 then you divide 2 by both sides and you get x which is
x > -6

8 0

3 years ago

In a game, you toss a fair coin and a fair six-sided die. If you toss a heads on the coin and roll either a 3 or a 6 on the die,

pochemuha

Using probabilities, it is found that the expected profit of one round of this game is of $0.

A probability is the number of desired outcomes divided by the number of total outcomes.

One of the two sides of the coin are heads.
2 of the 6 sides of the dice are 3 or 6.

Hence, since the coin and the dice are independent, the probability of winning is:

$p = \frac{1}{2} \times \frac{2}{6} = \frac{1}{6}$

The expected value is the sum of each outcome multiplied by its respective probability.

In this problem:

$\frac{1}{6}$ probability of earning $30.
$\frac{5}{6}$ probability of losing $6.

Then:

$E(X) = 30\frac{1}{6} - 6\frac{5}{6} = 5 - 5 = 0$

The expected profit of one round of this game is of $0.

A similar problem is given at brainly.com/question/24855677

5 0

3 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of 553 bags per hour. (a-1) At α = .05 i

timurjin [86]

Answer:

H1 : μ < 553. Reject H1 if tcalc > –1.753.

There is not enough evidence to reject the manufacturer’s claim.

Step-by-step explanation:

We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 553 bags per hour.

A sample of 16 randomly chosen hours with a mean of 533 and a standard deviation of 47 is given.

Let $\mu$ = average bags that an airport baggage scanning machine can handle.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 553 bags {means that the manufacturer’s claim is not overstated}

Alternate Hypothesis, $H_0$ : $\mu$ < 553 bags {means that the manufacturer’s claim is overstated}

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

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

where, $\bar X$ = sample mean = 533 bags

s = sample standard deviation = 47

n = sample of hours = 16

So, the test statistics = $\frac{533-553}{\frac{47}{\sqrt{16} } }$ ~ $t_1_5$

= -1.702

The value of t test statistics is -1.702.

Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.702 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the manufacturer’s claim is not overstated and an airport baggage scanning machine can handle an average of 553 bags per hour.

5 0

3 years ago