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3 years ago

I need help fast please I need it quckily

Mathematics

1 answer:

MariettaO [177]3 years ago

7 0

Hi there!

First, let's figure out how many feet are in the total amount of tape. This can be done by multiplying the amount of yards (4) and multiplying it by the amount of feet in a yard (3). This gives us 12 feet. Then, we need to subtract the amount of feet that are left (2). This gives us 10 feet.

ANSWER:
B - 10 feet

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

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22 quarts of orange juice is available. How many gallons of orange juice is available?

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5.5 gallons

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3 years ago

Find the reference angle for 497° A) 137° B)133° C) 47° D) 43°

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3 years ago

Read 2 more answers

In order to ensure efficient usage of a server, it is necessary to estimate the mean number

juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± $Z_{1-\alpha /2}$ * S/√n]

$Z_{1-\alpha /2}$ = $Z_{0.95}$ = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

$Z_{1\alpha } = Z_{0.99} = 2.33$

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z= X[bar] - μ = 37.7 - 35 = 2.93

S/√n 9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

Note: To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.

I hope it helps!

7 0

3 years ago

A shipment of 50,000 transistors arrives at a manufacturing plant. The quality control engineer at the plant obtains a random sa

Aleks04 [339]

Step-by-step explanation:

remember, the number of possible combinations to pick m out of n elements is C(n, m) = n!/(m! × (n-m)!)

50,000 transistors.

4% are defective, that means 4/100 = 1/25 of the whole.

so, the probability for one picked transistor to be defective is 1/25.

and the probability for it to work properly is then 1-1/25 = 24/25.

now, 500 picks are done.

to accept the shipment, 9 or less of these 500 picks must be defective.

the probability is then the sum of the probabilities to get

0 defective = (24/25)⁵⁰⁰

1 defective = (24/25)⁴⁹⁹×1/25 × C(500, 1)

= 24⁴⁹⁹/25⁵⁰⁰ × 500

2 defective = (24/25)⁴⁹⁸×1/25² × C(500, 2)

= 24⁴⁹⁸/25⁵⁰⁰ × 250×499

3 defective = 24⁴⁹⁷/25⁵⁰⁰ × C(500, 3) =

= 24⁴⁹⁷/25⁵⁰⁰ × 250×499×166

...

9 defective = 24⁴⁹¹/25⁵⁰⁰ × C(500, 9) =

= 24⁴⁹¹/25⁵⁰⁰ × 500×499×498×497×496×495×494×493×492×491 /

9×8×7×6×5×4×3×2 =

= 24⁴⁹¹/25⁵⁰⁰ × 50×499×166×71×31×55×494×493×41×491

best to use Excel or another form of spreadsheet to calculate all this and add it all up :

the probability that the engineer will accept the shipment is

0.004376634...

which makes sense, when you think about it, because 10 defect units in the 500 is only 2%. and since the whole shipment contains 4% defect units, it is highly unlikely that the random sample of 500 will pick so overwhelmingly the good pieces.

is the acceptance policy good ?

that completely depends on the circumstances.

what was the requirement about max. faulty rate in the first place ? if it was 2%, then the engineer's approach is basically sound.

it then further depends what are the costs resulting from a faulty unit ? that depends again on when the defect is usually found (still in manufacturing, or already out there at the customer site, or somewhere in between) and how critical the product containing such transistors is. e.g. recalls for products are extremely costly, while simply sorting the bad transistors out during the manufacturing process can be rather cheap. if there is a reliable and quick process to do so.

so, depending on repair, outage and even penalty costs it might be even advisable to have a harder limit during the sample test.

in other words - it depends on experience and the found distribution/probability curve, standard deviation, costs involved and other factors to define the best criteria for the sample test.

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2 years ago

What is the solution to the equation log2 (5x - 2) = 3?

Advocard [28]

From log BNE to BEN

X=2

if you want explaination then ask me

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3 years ago