SOLVE 5 and 6 for me.

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$

$16\cdot t^{2}-90\cdot t +81 = 0$ (1)

By Quadratic Formula:

$t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}$

$t_{1} = 4.5\,s$ , $t_{2} = 1.125\,s$

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0

3 years ago

drug sniffing dogs must be 95% accurate in their responses because their handlers don't want them to miss durgs and also don't w

GenaCL600 [577]

Answer:

95% Confidence interval: (0.8449,0.9951)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 50

Number of times the dog is right, x = 46

$\hat{p} = \dfrac{x}{n} = \dfrac{46}{50} = 0.92$

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get:

$0.92 \pm 1.96(\sqrt{\dfrac{0.92(1-0.92)}{50}})\\\\ = 0.92\pm 0.0751\\\\=(0.8449,0.9951)$

(0.8449,0.9951) is the required 95% confidence interval for the proportion of times the dog will be correct.

7 0

3 years ago

Please help! A-level maths.

astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.