In a mathematics class, half of the students scored 84 on an achievement test. With the

The mean survivial time after diagnosis for a certain disease is 15 years with a standard deviation of 5 years. Based on a parti

frutty [35]

Answer:

The time the patient expected to survive after diagnosis is 29 years.

Step-by-step explanation:

It is provided that the mean survival time after diagnosis for a certain disease is 15 years with a standard deviation of 5 years.

That is,

$\mu=15\\\sigma=5$

An individual's predicted survival time is <em>a</em> = 2.8 standard deviations beyond the mean.

Compute the time the patient expected to survive after diagnosis as follows:

$X=\mu+a\sigma$

$=15+(2.8\times 5)\\\\=15+14\\\\=29$

Thus, the time the patient expected to survive after diagnosis is 29 years.

5 0

3 years ago

How do I solve -2x > 6

vodomira [7]

I’m not the best with words but first you would have to treat the greater sign as a = sign. So..
-2x =6
/ /
-2 -2
X = -3
and then you would change the sign to a less than sign because you divided by a negative.

7 0

3 years ago

There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am

Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

$n = 479, \pi = \frac{130}{479} = 0.2714$

Hence the bounds of the interval are found as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112$

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0

2 years ago

Can someone please help me on this!!

pashok25 [27]

Answer:

(Look at image)

Step-by-step explanation:

(Also look at image)

5 0

3 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago