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2 months ago

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Sam receives the following scores on his English tests: 63, 84, and 96. What average score does he need on the last two tests in

order to maintain an 85 average?

1 answer:

Zina [12.3K]2 months ago

4 0

Sam needs to score 97 on his upcoming test to keep his average at 85. If you total 63, 84, 96, and 97, the sum is 340. Dividing 340 by four test scores yields an exact average of 85.

You might be interested in

The idle time for taxi drivers in a day are normally distributed with an unknown population mean and standard deviation. If a ra

Svet_ta [12734]

Answer:

$172-2.51\frac{16}{\sqrt{23}}=163.626$

$172+2.51\frac{16}{\sqrt{23}}=180.374$

Hence, in this case, the 98% confidence interval would be (163.626;180.374)

Step-by-step breakdown:

Previous concepts

A confidence interval represents a range that is likely to encompass a population value within a specific confidence level, typically expressed as a percentage whereby a population mean falls between an upper and lower limit.

The margin of errorindicates the span of values surrounding the sample statistic in a confidence interval.

A normal distributionillustrates a probability distribution that is symmetrical around the mean, signifying that values near the mean occur more frequently than those farther away from it.

$\bar X=172$ denote the sample mean

$\mu$ population mean (the variable of interest)

s=16 signifies the sample standard deviation

n=23 represents the sample size

The solution to the query

The equation for the confidence interval of the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

To determine the critical value $t_{\alpha/2}$ , we first need to calculate the degrees of freedom, which is expressed as:

$df=n-1=23-1=22$

Since the confidence level is 0.98 or 98%, we find the value of $\alpha=0.02$ and $\alpha/2 =0.01$ using tools like Excel or a calculator, where the Excel command would be: "=-T.INV(0.01,22)". This yields $t_{\alpha/2}=2.51$

Having all components ready, we can substitute into formula (1):

$172-2.51\frac{16}{\sqrt{23}}=163.626$

$172+2.51\frac{16}{\sqrt{23}}=180.374$

Thus, for this case, the 98% confidence interval will be (163.626;180.374)

3 0

1 month ago

Maia had 2064 more beads than Jenny. After Maia used 144 beads to make a necklace, she had 5 times as many beads as Jenny. A) ho

tester [12383]

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:

$y=x+2064...(1)$

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:

$y-144=5x...(2)$

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

$\text{Number of beads Maia had more than Jenny in the end}=2064-144$

$\text{Number of beads Maia had more than Jenny in the end}=1920$

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:

$x+2064-144=5x$

$x+1920=5x$

$x-x+1920=5x-x$

$1920=4x$

Now, let's divide our equation by 4.

$\frac{1920}{4}=\frac{4x}{4}$

$480=x$

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

$y=480+2064$

$y=2544$

Conclusively, Maia initially had 2544 beads.

3 0

2 months ago