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2 years ago

What is the following product 3 sqrt 2(5 sqrt 6-7 sqrt 3)

Mathematics

2 answers:

harkovskaia [24]2 years ago

8 0

The answer in the attached figure

inysia [295]2 years ago

6 0

Answer:

$(30\sqrt{3}-21\sqrt{6})$

Step-by-step explanation:

We have to find the product of the expressions.

$3\sqrt{2}(5\sqrt{6}-7\sqrt{3})\\$

= $3\sqrt{2}\times 5\sqrt{6}-7\sqrt{3}\times 3\sqrt{2}$ ( By distributive rule)

= $3\times 5(\sqrt{2})(\sqrt{6})-7\times 3(\sqrt{3})(\sqrt{2})$

= $15\sqrt{12}-21\sqrt{6}$

= $15\sqrt{4\times 3}-21\sqrt{6}$

= $30\sqrt{3}-21\sqrt{6}$

= $(30\sqrt{3}-21\sqrt{6})$

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A population has a standard deviation of 80. A random sample of 400 items from this population is selected. The sample mean is d

Helga [31]

Answer:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =80$ represent the population standard deviation

n=400 represent the sample size

Solution to the problem

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

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

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.8808$

The margin of error is given by:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$

4 0

2 years ago

Trapezoid STUV with vertices S(-3,6),T(0,7),U(1,4),and V(-5,2): (x,y)>(x+7,y-9)

hjlf

Answer:

See below

Step-by-step explanation:

Given points:

S(-3,6),T(0,7),U(1,4),and V(-5,2)

Translation rule:

(x,y) → (x+7,y-9)

New coordinates by applying the rule:

S'(4, -3), T'(7, -2), U'(8, -5), V'(2, -7)

See attached for the graph

The blue trapezoid is STUV and the red one is S'T'U'V'

7 0

2 years ago

The sales data for January and February of a frozen yogurt shop are approximately normal. The mean daily sales for January was $

VashaNatasha [74]

Answer:

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

Step-by-step explanation:

z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

January:

The mean daily sales for January was $300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. This means, respectively, that $\mu = 300, \sigma = 20, X = 310$ . So

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

$Z = \frac{310 - 300}{20}$

$Z = 0.5$

February:

The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. This means, respectively, that $\mu = 320, \sigma = 50, X = 340$ . So