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3 years ago

8

To celebrate the first day of school, Austin brought in a tray of caramel brownies and walnut brownies. He brought 14 caramel br

ownies and 6 walnut brownies. What percentage of the brownies were caramel?

1 answer:

kramer3 years ago

5 0

The answer is 70%. There were 20 brownies total (14+6), and 14 out of 20 were caramel. 14/20 as a percent is 70%!

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A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviatio

ss7ja [257]

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

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5 0

3 years ago

I need my smart homies to answer this question for me !

ella [17]

Answer:

Part A: chosen method is by factoring

Part B: First rewrite the equation by writing -24 as a difference so you get x² -9x - 15x + 135 = 0. Then factor out the x and the -15 from the equation so you get x(x-9) - 15(x-9) = 0. Then factor out x-9 from the equation so you have (x-9)(x-15) = 0. Finally, set each expression to 0: x-9=0 and x-15=0 and then solve: x=9 and x=15

Part C: x=9, x=15

4 0

3 years ago

Help! ill give points

professor190 [17]

Answer:

108

Step-by-step explanation:

5 0

2 years ago

A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib

Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 502, \sigma = 115$

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{590 - 502}{115}$

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{400 - 502}{115}$

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

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6 0

2 years ago

Don’t know the answer to this?

Vaselesa [24]

Answer:

Step-by-step explanation:

The first one.

3 0

3 years ago