Ask question

Ask question

All categories

3 years ago

13

A certain tennis player makes a successful first serve 70% of the time. Assume that each serve is independent of the others. If

she
serves 6 times, what's the probability that she gets...
All 6 serves in?
Exactly 4 serves in?
At least 4 serves in?
No more than 4 serves in?

1 answer:

EleoNora [17]3 years ago

6 0

this is a binomial problem: p = 0.7 and q = 0.3

a) (0.7)^6

b) (6C4)(0.7)^4(0.3)^2

c) Pr ( at least 4) = Pr(4) + Pr(5) + Pr(6) = (6C5)(0.7)^5(0.3) + (0.7)^6

d) Pr (no more than 4) = 1 - Pr(at least 4) = 1 - (answer from c)

You might be interested in

It is given that y varies inversely as x^2. If x is decreased by 50%, find the percentage change in y.

vova2212 [387]

Answer:

300%

Step-by-step explanation:

Given

$y\ \alpha\ \frac{1}{x^2}$

Required

Find the percentage change in y when x decreased by 50%

First, convert to equation

$y\ = \ \frac{k}{x^2}$

Where k is the constant of proportionality

When x decreased by 50%

$Y = \frac{k}{(x - 0.5x)^2}$

$Y = \frac{k}{(0.5x)^2}$

$Y = \frac{k}{0.25x^2}$

Expand

$Y = \frac{1}{0.25} * \frac{k}{x^2}$

Substitute $\frac{k}{x^2}$ for y

$Y = \frac{1}{0.25} * y$

$Y = 4 * y$

$Y = 4 y$

The percentage change is then calculated as:

$\%Change = \frac{Y - y}{y} * 100\%$

$\%Change = \frac{4y - y}{y} * 100\%$

$\%Change = \frac{3y}{y} * 100\%$

$\%Change = 3 * 100\%$

$\%Change = 300\%$

<em>The percentage in y is 300%</em>

7 0

3 years ago

PLS HELP ME AND SHOW WORK.

Alex17521 [72]

Answer:

$4 = (x + 7 {)}^{2} + (y - 3 {)}^{2}$

Step-by-step explanation:

equation of a circle

${r}^{2} = (x - {h} )^{2} + (y - {k)}^{2}$

we know the center is (-7, 3)

the radius is 2

we replace (h,k) with the center coordinates

and input the radius

${2}^{2} = (x -( - 7)) ^{2} + (y - 3)^{2}$

simplify to this:

$4 = (x + 7 {)}^{2} + (y - 3 {)}^{2}$

3 0

3 years ago

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

sergeinik [125]

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$ , so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.

5 0

3 years ago

Helpppppp meeeeeeeee

seropon [69]

Answer:

its B im pretty sure.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

I really need the answer rnn so plss helpp

3241004551 [841]

Answer: Each shape has 12 edges

Step-by-step explanation:

7 0

3 years ago