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3 years ago

What is the probability that a 44% free throw shooter will miss her next free throw?

1 answer:

3 0

The probability is 56/100 or 56%. Because they only got 44% of their previous throws, the probability that they'll miss the next one is 56%.

You might be interested in

The following graph shows the cost c, in dollars, of p pounds of strawberries at a local grocery store.

lorasvet [3.4K]

Answer:

A. The ordered pair (5, 3.5) represents 5 lbs. of strawberries cost $3.50.

B. The cost per pound of strawberries is shown at (1, 0.7).

C. The relationship shown in the graph is a direct variation because as the pounds of strawberries increase, the cost also increases.

Explanation:

A. The coordinate is in (x,y). The x-axis represents the pound of strawberries and the y-axis represents the cost of the strawberries.

B. The cost for one pound of strawberries can be calculated to find the exact value of y (cost) when x is 1. The graph shows 5 lbs. of strawberries cost $3.50 so divide $3.50 by 5 lbs. to find the cost per pound of strawberries. 3.5 ÷ 5 = 0.7. On the graph, the cost for one pound of strawberries is at (1, 0.7)

C. A direct variation in a graph is shown when both x and y increase or decrease together. In this case, as the pounds of strawberries increase, the cost of the strawberries also increases.

6 0

3 years ago

The value of a car is inversely proportional to its age. if a car is worth $8100 when it is 6 years old, how old will it be wh

Sveta_85 [38]

It is $13.500 because you are adding

8 0

2 years ago

The weight of an organ in adult males has a bell-shaped distribution with a mean of 320 grams and a standard deviation of 45 gr

irina1246 [14]

Answer:

a) 99.7%

b) 68%

c) 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

d) 81.5% of organs weighs between 290 grams and 365 grams.

Step-by-step explanation:

A) 99.7% of organs will be between 3 standard deviation from the mean.

$320 - 3 \times 45 = 185$

$320 + 3 \times 45 = 455$

So 99.7% of organs will be between 185 and 455.

B) 275 grams and 365 grams are 1 standard deviation from the mean.

From empirical rule about 68% data falls within 1 standard deviation from the mean.

So 68% of organs weighs btwn 275 grams and 365 grams.

C) Since 68% of organs weighs between 275 and 365 grams, so 32% of organs weighs will be less than 275 or more than 365 grams.

D) 230 is 2 standard deviation below the mean and 365 is one standard deviation above the mean.

According to the empirical rule, 95% of the observation lies within 2 standard deviations of the mean. Therefore 5% lies outside 2 standard deviations of the mean.

So 95%/2 = 47.5% of organs weighs between the mean and 230 grams.

According to the empirical rule about 68% data falls within one standard deviation from the mean.

so, 68%/2 = 34% of organs weighs between the mean and 365 grams.

So total = 47.5% + 34% = 81.5% of organs weighs between 290 grams and 365 grams.

7 0

3 years ago

Add or subtract as indicated and write the result in standard form. 5i - (-5 - i)

AlexFokin [52]

Simplify by combing the real and imaginary parts of each expression.

5 + 6i

Arrange the expression into the a+bi form of a complex number.

5+ 6i

5 0

2 years ago

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi

Luden [163]

Answer:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size

$\mu_o =3.3$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

6 0

2 years ago