Madeline usually makes 85% of her shots in basketball. if she attempts 20, how many will she likely make?​

Answer:

A) Maia had 1920 beads more.

B) Maia had 2544 beads at first.

Step-by-step analysis:

Let x denote the beads with Jenny and y for Maia.

The information provided states that Maia possesses 2064 beads more than Jenny, which can be represented mathematically as:

Additionally, after Maia made a necklace with 144 beads, she had five times more beads than Jenny.

This can also be formulated as an equation:

A) Since Maia had 2064 beads more than Jenny before using 144 beads, we calculate her final bead count by subtracting 144 from 2064.

Thus, Maia is left with 1920 beads more than Jenny.

B) To solve this system of linear equations, we will utilize the substitution method.

By inserting equation (1) into equation (2), we arrive at:

Now, let's divide our equation by 4.

Now, we'll substitute x=480 into equation (1) to isolate y.

Conclusively, Maia initially had 2544 beads.

Answer: 10 holly bushes and 12 bayberry shrubs

Step-by-step explanation: use your brain or a calculator to figure out the cost of each type of plant and then total them up

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

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Answer:

13.8%

Step-by-step explanation:

With six employees and six checks, there are a total of 36 (6x6) possible distributions. To determine the probability that exactly five individuals receive the correct checks, there’s only one correct distribution for each of them out of the 36 possibilities, resulting in the calculation of 1/36 for the first and then 1/36 for the subsequent checks. Therefore, the final probability becomes 5/36 = 13.8%.

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