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kobusy [5.1K]
2 years ago
11

Help me i actually need this asap...Only right answer ..​No spammer.No wrong answer.

Mathematics
2 answers:
Leona [35]2 years ago
7 0
  1. \sqrt[3]{8} \: -  \sqrt{3}
katrin2010 [14]2 years ago
6 0

.............5b............

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ludmilkaskok [199]
-6 is the right answer.
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3 years ago
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( 20points please ) -22+(-16)=
hjlf

-22+(-16)

-22-16= -38

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3 years ago
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Rose and Andrew are financing 128,000 to purchase a condominium. They obtained a 15 year fixed rate loan with a rate of 5.05%. T
anzhelika [568]

Amount Financed(P) = $ 128,000

Time = 15 year

Rate of interest= 5.05%

Amount after 15 years translates to,

{\text{Using the formula}} P(1+\frac{R}{100})^n\\\\ 128,000(1 +\frac{5.05}{100})^{15}\\\\ =128,000 \times (1.0505)^{15}\\\\=128,000 \times 2.0938\\\\=268,009.890

Rose and Andrew has taken four points to lower their rate to 4.81% .

   Amount after 15 years translates to, if rate of interest is 4.83 %

{\text{Using the formula}} P(1+\frac{R}{100})^n\\\\ 128,000(1 +\frac{4.83}{100})^{15}\\\\ =128,000 \times (1.0483)^{15}\\\\=128,000 \times 2.0290\\\\=259713.0404

Cost of four points = 268,009.890 - 259,713.0404

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3 0
2 years ago
A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m
love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is c = 20.6153.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that \mu = 20.4, \sigma = 1.23

Consider the sample mean weight of 100 watermelons of this variety. This means that n = 100.

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when X = 20.27.

Since we are working with the sample mean, we use s instead of \sigma in the Z score formula

Z = \frac{X - \mu}{s}

Z = \frac{20.27 - 20.4}{0.123}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of X = c that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use Z = 1.75

Z = \frac{X - \mu}{s}

1.75 = \frac{c - 20.4}{0.123}

c - 20.4 = 0.123*1.75

c = 20.6153

This value is c = 20.6153.

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Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. x2 – 6 = x
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Answer:

3, -2

Step-by-step explanation:

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