The new playground equipment for the park was delivered in the box shown.

Bulls records has the same sensor when is the area of the shaded region

GrogVix [38]

Answer:

$2835.42mi^2$

Step-by-step explanation:

First things first, we would need to find the area of both circles.

Let's find the blue one first.

Use the circle area formula.

$A=\pi r^2$

$A=3.14r^2$

Now we need to find the radius of the circle.

Since they have the same center, we can add up the radius of the white circle and the line from the blue circle.

The radius of the white circle is half of the diameter so the radius is 11.

So the whole radius is 11 + 21 or 32.

Plug in 32 for $r$ .

$A=3.14(32)^2 \\ A=3.14(1024)\\A=3215.36$

So the area of the blue circle is 3215.36.

Now we need to find the area of the white circle.

Use the same formula and radius found above.

$A=3.14(11)^2 \\ A=3.14(121) \\ A=379.94$

So the area of the white circle is 379.94.

To find the area of the shaded region, all we have to do is subtract the area of the white circle from the area of the blue circle.

$3215.35-379.94=2835.42$

6 0

3 years ago

An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It descr

Step2247 [10]

Answer:

We conclude that the average attention span for teenage boys is less than 5 minutes.

Step-by-step explanation:

We are given that the mean time to distraction for teenage boys working on an independent task was 4 minutes.

Suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was 1.4 minutes.

Let $\mu$ = average attention span for teenage boys

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 5 minutes {means that the average attention span for teenage boys is more than or equal to 5 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 5 minutes {means that the average attention span for teenage boys is less than 5 minutes}

The test statistics that will be used here is 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 attention time span for teenage boys = 4 min

s = sample standard deviation = 1.4 min

n = sample of teenage boys = 50

So, the test statistics = $\frac{4-5}{\frac{1.4}{\sqrt{50} } }$ ~ $t_4_9$

= -5.051

Now at 0.01 significance level, the t table gives critical value of -2.405 at 49 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -2.405 > -5.051, 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 average attention span for teenage boys is less than 5 minutes.

8 0

4 years ago

Find a number that is 15 more than 4 times its opposite

svetlana [45]

X - the number

$x=4\cdot(-x)+15\\
x=-4x+15\\
x+4x=15\\
5x=15\\
x=3$