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2 years ago

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NO ONE WILL HELP IN ENGLISH CAN SOMEONE HELP HERE!! GIVING AWAY 20 PNTS AND GIVING BRAINLIEST TO THE BEST ANSWER!!!!!!!!!!!

1 answer:

olchik [2.2K]2 years ago

6 0

Are these true and false? I think it's true for both

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In a standard deck of cards there are four suits: spades, clubs, hearts, and diamonds. Each suit has one each of 13 cards: ace,

olasank [31]

Answer:

The answer would be graph A.

Step-by-step explanation:

Since the cards all belong to four suites, the king can be a heart as well as the jack, but they cannot be the same card. So the choice is graph A.

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3 years ago

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NEED UR HELPPPP PLEASE..

Mashcka [7]

17.32 deal with it aaaaaaaaaaaaaaaaa

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The cost of a week long vacation in a city has a mean of 1000 and a variance of 1200. The city imposes a tax which will raise al

matrenka [14]

Answer:

The coefficient of variation after the tax is imposed is 0.033

Step-by-step explanation:

Given

$\mu =1000$ --- mean

$\sigma^2 = 1200$ --- variance

$tax = 10\%$

Required

The coefficient of variation

The coefficient of variation is calculated using:

$CV = \frac{\sqrt{\sigma^2}}{\mu}$

After the tax, the new mean is:

$\mu_{new} = \mu * (1 + tax)$

$\mu_{new} = 1000 * (1 + 10\%)$

$\mu_{new} = 1100$

And the new variance is:

$\sigma^2_{new} = \sigma^2 * (1 + tax)$

$\sigma^2_{new} = 1200 * (1 + 10\%)$

$\sigma^2_{new} = 1320$

So, we have:

$CV = \frac{\sqrt{\sigma^2}}{\mu}$

$CV = \frac{\sqrt{1320}}{1100}$

$CV = \frac{36.33}{1100}$

$CV = 0.033$

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2 years ago

What is the equation of the line shown in this graph?

dolphi86 [110]

X=-2 is shown on the graph.

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2 years ago

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A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From

lutik1710 [3]

Answer:

a) $z = 1.645$

b) The should sample at least 293 small claims.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ , which means that the answer of question a is z = 1.645.

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

$M = 12, \sigma = 124.88$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$12 = 1.645*\frac{124.88}{\sqrt{n}}$

$12\sqrt{n} = 205.43$

$\sqrt{n} = \frac{205.43}{12}$

$\sqrt{n} = 17.12$

$\sqrt{n}^{2} = (17.12)^{2}$

$n = 293$

The should sample at least 293 small claims.

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2 years ago