I need help asap will give brainiest

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Answer 1

Answer:

Answer:

Step-by-step explanation:

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Will wanted to track the growth of various fruits in his garden, so he decided to label them. His garden had APPLES labeled 1,2,3,4,5,6, LEMONS labeled 1,2, and MELONS labeled 1,2,3. If a single fruit is picked at random, what is the probability that the fruit is an APPLE or has an EVEN number?

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Using the probability concept, it is found that there is a 0.7273 = 72.73% probability that the fruit is an APPLE or has an EVEN number.

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem:

There are 11 options, hence $T = 11$ .
Of those, there are 6 options which land on Apple, plus 2 options which does not land on apple but have an even number, hence $D = 6 + 2 = 8$

Then, the probability is:

$p = (D)/(T) = (8)/(11) = 0.7273$

0.7273 = 72.73% probability that the fruit is an APPLE or has an EVEN number.

A similar problem involving the probability concept is given at brainly.com/question/15536019

Answer:

8 in 11 or 0.7272

Step-by-step explanation:

Between 6 apples. 2 lemons and 3 melons, Will has a total of 11 fruits in his garden. There are three apples labeled with even numbers (2,4 and 6), one melon (2) and one lemon (2), for a total of five fruits.

The probability that a randomly selected fruit is an APPLE or has an EVEN number is given by the probability that it is an apple, P(A), added to the probability that it is even, P(E), minus the probability that it is an even apple, P(A and E):

$P(A\ or\ E) = P(A) +P(E) - P(A\ and\ E)\nP(A\ or\ E) = (6)/(11)+(5)/(11)-(3)/(11) \nP(A\ or\ E) = 0.7272$

There is a 8 in 11, or a 0.7272 chance that the fruit is an APPLE or has an EVEN number.

Blood potassium level, continued. Judy’s measured potassium level varies according to the Normal distribution with with μ=3.8 and σ=0.2mmol/l. Let’s consider what could happen if we took 4 separate measurements from Judy. What is the blood potassium level L such that the probability is only 0.05 that the average of 4 measurements is less than L? (Hint: This requires a backward Normal calculation.)the book says about 3.64 but i got 3.28

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Answer:

The value of l = 3.64

Step-by-step explanation:

Consider the provided information.

μ=3.8 and σ=0.2mmol/l and n=4

As we know $\mu_(\bar x)=\mu=3.8$

$\sigma_(\bar x)=(\sigma)/(√(n))=(0.2)/(√(4))=0.1$

We have given α = 0.05

So by the standard normal table: $Z_(0.05)=-1.645$

Compute $P(\bar{X}<l)=0.05$

$P(z<(l-3.8)/(0.1))=0.05$

$(l-3.8)/(0.1)=-1.645$

$l-3.8=-0.1645$

$l=3.64$

Hence, the value of l = 3.64

Which of the following groups have terms that can be used interchangeably?a. critical value, probability, proportion
b. percentage, probability, proportion
c. critical value, percentage, proportion
d. critical value, percentage, probability

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percentage, probability, proportion are all very similar terms (not 100% identical but closely related)

The answer is choice B

Which of the following constants can be added to x2 + x to form a perfect square trinomial?

Answers

Answer:

x^2+x + 1/4

Step-by-step explanation:

x^2+x

Take the coefficient of x

divide by 2

1/2

Square it

(1/2)^2 = 1/4

Add this to make a perfect square trinomial

The Graduate Record Examination (GRE) is a standardized test that students usually take before entering graduate school. According to the document Interpreting Your GRE Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE are (approximately) normally distributed with mean 462 points and standard deviation 119 points. (6 p.) (a) Obtain and interpret the quartiles for these scores. (b) Find and interpret the 99th percentile for these scores

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Answer:

(a) The first quartile is 382.27 and it means that at least el 25% of the scores are less than 382.27 points.

The second quartile is 462 and it means that at least el 50% of the scores are less than 462 points.

The third quartile is 541.73 and it means that at least el 75% of the scores are less than 541.73 points.

(b) The 99th percentile is 739.27 and it means that at least el 99% of the scores are less than 739.27 points.

Step-by-step explanation:

The first, second the third quartile are the values that let a probability of 0.25, 0.5 and 0.75 on the left tail respectively.

So, to find the first quartile, we need to find the z-score for which:

P(Z<z) = 0.25

using the normal table, z is equal to: -0.67

So, the value x equal to the first quartile is:

$z=(x-m)/(s)\n x=z*s +m\nx =-0.67*119 + 462\nx=382.27$

Then, the first quartile is 382.27 and it means that at least el 25% of the scores are less than 382.27 points.

At the same way, the z-score for the second quartile is 0, so:

$x=0*119+462\nx=462$

So, the second quartile is 462 and it means that at least el 50% of the scores are less than 462 points.

Finally, the z-score for the third quartile is 0.67, so:

$x=z*s +m\nx =0.67*119 + 462\nx=541.73$

So, the third quartile is 541.73 and it means that at least el 75% of the scores are less than 541.73 points.

Additionally, the z-score for the 99th percentile is the z-score for which:

P(Z<z) = 0.99

z = 2.33

So, the 99th percentile is calculated as:

$x=z*s +m\nx =2.33*119 + 462\nx=739.27$

So, the 99th percentile is 739.27 and it means that at least el 99% of the scores are less than 739.27 points.

A direct mailing company sells computers and computer parts by mail. The company claims that at least 90% of all orders are mailed within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid. A recently taken random sample of 150 orders showed that 129 of them were mailed within 72 hours. Set up the null and alternative hypotheses to test whether the company’s claim is true?

Answers

Answer:

The null and alternative hypothesis are:

$H_0: \pi=0.9\n\nH_a:\pi<0.9$

There is not enough evidence to support the claim that the company is worng and less than 90% of all orders are mailed within 72 hours after they are received.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The company claims that at least 90% of all orders are mailed within 72 hours after they are received. The company's claim is the null hypothesis, the claim that is going to be tested is if this proportion is significantly less than 90%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.9\n\nH_a:\pi<0.9$

The significance level is 0.05.

The sample has a size n=150.

The sample proportion is p=0.86.

$p=X/n=129/150=0.86$

The standard error of the proportion is:

$\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.9*0.1)/(150)}\n\n\n \sigma_p=√(0.0006)=0.024$

Then, we can calculate the z-statistic as:

$z=(p-\pi+0.5/n)/(\sigma_p)=(0.86-0.9+0.5/150)/(0.024)=(-0.037)/(0.024)=-1.497$

This test is a left-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z<-1.497)=0.0672

As the P-value (0.0672) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that less than 90% of all orders are mailed within 72 hours after they are received.

Final answer:

The null hypothesis is that 90% of all orders are shipped within 72 hours, denoted as H0: p = 0.90. The alternative hypothesis is that less than 90% of all orders are shipped within 72 hours, expressed as HA: p < 0.90.

Explanation:

In hypothesis testing, the null hypothesis represents the theory that has been put forward, either because it is believed to be true or because it is used as a basis for argument. In this case, the null hypothesis would be that 90% of all orders are shipped within 72 hours. This can be formulated as H0: p = 0.90 where p represents the proportion of all orders shipped within 72 hours.

The alternative hypothesis on the other hand is the hypothesis used in hypothesis testing that is contrary to the null hypothesis. Here, the alternative hypothesis would be that less than 90% of all orders are shipped within 72 hours. It can be expressed as HA: p < 0.90 where p remains the proportion of orders shipped within 72 hours.

Learn more about Hypothesis Testing here:

brainly.com/question/31665727

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In all, 54 children were assigned to the Montessori school and 115 to other schools at age three. When the children were five, parents of 38 of the Montessori children and 34 of the others could be located and agreed to and subsequently participated in testing. This information reveals a possible source of bias in the comparison of outcomes. Explain why. (Round your answers to the nearest whole number.) About % of Montessori parents participated in the study, compared to about % of the other parents. (c) One of the many response variables was score on a test of ability to apply basic mathematics to solve problems. Here are summaries for the children who took this test: