If you have a 40% chance of making a free throw, what is the probability of missing a free throw?

jack wants to weigh at most 176 pounds before the baseball season begins. he has already lost 15 pounds. he belives he can lose

VikaD [51]

The answer will be 4 weeks. You know the constant is 1.5 lbs per week and you will multiply that by however many weeks it takes to reach 176
197-(1.5x) would be your equation and since he’s already lost 15 pounds you will include that by subtracting 15 from the equation as well 197-(1.5(4))=191
191-15=176

6 0

1 year ago

Joan is hiking. She wants to cover 1,600 feet. She hikes 600 feet every 3 hours.

Stella [2.4K]

Joan's remaining distance is reduced by (600 ft)/(3 hours) = 200 ft/hour. She starts with 1600 ft remaining, so her distance remaining (y) after x hours is
.. y = -200x +1600

In order for the distance remaining to be zero, you must have
.. 0 = -200x +1600
.. 200x = 1600
.. x = 1600/200 = 8

It will take Joan 8 hours to hike 1600 ft.

6 0

1 year ago

A sample of size 40 is taken from an infinite population whose mean and standard deviation are 68 and 12 respectively. the proba

gulaghasi [49]

To solve this problem, we can use the z statistic to find for the probability. The formula for z score is:

z = (x – u) / s

Where,

u = the sample mean = 68

s = standard deviation of the samples = 12

x = sample value = 70

In this case, what is asked is the probability that the sample mean is larger than 70, therefore this corresponds to the right of z on the graph of normal distribution.

z = (70 – 68) / 12

z = 2 / 12

z = 0.17

Therefore finding for P at z ≥ 0.17 using the standard distribution tables:

P (z = 0.17) = 0.5675

But this is not the answer yet since this is the P to the left of z. Therefore the correct one is:

P (z ≥ 0.17) = 1 - 0.5675

P (z ≥ 0.17) = 0.4325 = 43.25%

<span>The probability that the sample mean is larger than 70 is 43.25%</span>

3 0

1 year ago

According to Money magazine, Maryland had the highest median annual household income of any state in 2018 at $75,847.† Assume th

Shtirlitz [24]

Answer:

a. 0.3372 or 33.72%; b. 0.2236 or 22.36%; c. 0.2879 or 28.79%; d. $112,351.00

Step-by-step explanation:

We need to use here the values from the <em>cumulative standard normal table</em> and <em>z-scores</em> to solve the questions. That is, all the values are transformed to a <em>z-score</em> to use the <em>standard normal table</em> to find the probabilities. Notice that the question is telling us about the <em>median</em> and not <em>mean</em>. Fortunately, in the normal distribution the mean, the median, and the mode are the same. So, we can say that:

$\\ median\;,mode\;,and\;mean=\mu$

As a result, the parameters for the normal distribution in this case are:

$\\ \mu = 75847\;and\;\sigma=33800$

Then we can solve the questions as follows:

<h3>Part a: Probability that annual income of 90,000 or more</h3>

We need to calculate the z-score of such a value of x=90,000:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{90000 - 75847}{33800}$

$\\ z = \frac{14153}{33800}$

$\\ z = 0.41872$

We need to round this value to z = 0.42 to use the <em>cumulative standard normal table. </em>This value is above the mean (positive) and corresponds, approximately, with a cumulative probability of P(x<90000) = 0.66276.

Then, the probability that a household in Maryland has an annual income of $90,000 or more is:

$\\ P(x\geq90000) = 1 - P(x$

Rounding to four decimal places is 0.3372 or 33.72%

<em>We can follow the same procedure to find the rest of the probabilities asked.</em>

<h3>Part b: Probability a household has an annual income of 50,000 or less.</h3>

$\\ P(x\leq50000) = \?$

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{50000 - 75847}{33800}$

$\\ z = -0.764704$

Rounding this value to two decimals (z = -0.76), we can conclude that this value is below the mean. To find this probability from the cumulative standard normal table, we first found the value of z = 0.76 (since no negative value is displayed in this table) and then subtracting this value from one. This is possible because the normal distribution is symmetrical. Then,

$\\ P(z$