What the answer to lesson 5.4

If the probability of surviving a head on car accident at 55 mph is 0.097, then what is the probability of not surviving

BabaBlast [244]

The probability of not surviving a head-on car accident is: 0.903

Step-by-step explanation:

The probabilities of occurrence and non-occurrence of an event add up to one.

If p is the probability that an even will happen and q is the probability that it will not happen

Then

$p+q=1$

Here,

Probability of surviving = p = 0.097

Probability of not surviving = q = ?

$0.097 + q = 1\\q = 1 - 0.097\\q = 0.903$

The probability of not surviving a head-on car accident is: 0.903

Keywords: Probability, Inverse

Learn more about probability at:

#LearnwithBrainly

5 0

3 years ago

a small rocket is shot from the edge of a cliff suppose that after t seconds the rocket is y meters above the cliff where y=30t-

Natalka [10]

Answer:

Greatest height: 45 meters

Time for greatest height: 3 seconds

Height after 5 seconds: 25 meters above the cliff

Time for height of 40 meters: 7.123 seconds

Height after 7 seconds: -35 meters (35 meters below the cliff)

Step-by-step explanation:

to find the maximum height, we need to calculate the derivative of y in relation to t and then find when dy/dt = 0:

dy/dt = 30 - 10t = 0

10t = 30

t = 3 seconds

In this time, the height is:

y = 30*3 - 5*3^2 = 45 meters

After 5 seconds, the height is:

y = 30*5 - 5*5^2 = 25 meters

The time for the height of 40 meters is:

40 = 30t - 5t^2

t^2 - 6t - 8 = 0

Using Bhaskara's formula, we have:

Delta = 6^2 + 4*8 = 68

sqrt(Delta) = 8.246

t1 = (6 + 8.246) / 2 = 7.123 seconds

t2 = (6 - 8.246) / 2 = -1.123 seconds (negative value for time is not valid)

So the time when the rocket reaches 40 meters is 7.123 seconds

After 7 seconds, the height is:

y = 30*7 - 5*7^2 = -35 meters

The rocket will be 35 meters below the cliff.

5 0

3 years ago

Would the answer for 36 be $9.88 or $9.75?

ludmilkaskok [199]

total bill before any discounts $13.

she is a student, so she gets 20% off.

she brought an item of clothing, so she gets 5% off.

so she's really getting 20% + 5% off, namely 25% off her bill.

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{25\% of 13}}{\left( \cfrac{25}{100} \right)13\implies 3.25}~\hfill \stackrel{\textit{bill with all discounts}}{13-3.25\implies 9.75}$

6 0

3 years ago

The p-value for a one-sided test of the population proportion is 0.0513. Which of

uysha [10]

Using the equation of the test statistic, it is found that with an increased sample size, the test statistic would decrease and the p-value would increase.

<h3>How to find the p-value of a test?</h3>

It depends on the test statistic z, as follows.

For a left-tailed test, it is the area under the normal curve to the left of z, which is the <u>p-value of z</u>.

For a right-tailed test, it is the area under the normal curve to the right of z, which is <u>1 subtracted by the p-value of z</u>.

For a two-tailed test, it is the area under the normal curve to the left of -z combined with the area to the right of z, hence it is <u>2 multiplied by 1 subtracted by the p-value of z</u>.

In all cases, a higher test statistic leads to a lower p-value, and vice-versa.

<h3>What is the equation for the test statistic?</h3>

The equation is given by:

$t = \frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{X}$ is the sample mean.

$\mu$ is the tested value.

s is the standard deviation.

n is the sample size.

From this, it is taken that if the sample size was increased with all other parameters remaining the same, the test statistic would decrease, and the p-value would increase.

You can learn more about p-values at brainly.com/question/26454209

3 0

2 years ago

The plans for a new ammusement park were laid out on the first quadrant of a coordinate plane. The entrance to the roller coaste

Solnce55 [7]

The first step is to determine the distance between the points, (1,1) and (7,9)

We would find this distance by applying the formula shown below

$\begin{gathered} \text{Distance = }\sqrt[]{(x2-x1)^2+(y2-y1)^2} \\ \text{From the graph, } \\ x1\text{ = 1, y1 = 1} \\ x2\text{ = 7, y2 = 9} \\ \text{Distance = }\sqrt[]{(7-1)^2+(9-1)^2} \\ \text{Distance = }\sqrt[]{6^2+8^2}\text{ = }\sqrt[]{100} \\ \text{Distance = 10} \end{gathered}$

Distance = 10 units

If one unit is 70 meters, then the distance between both entrances is

70 * 10 = 700 meters

4 0

1 year ago