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3 years ago

Last question for today.

Mathematics

2 answers:

djyliett [7]3 years ago

7 0

Answer:

Step-by-step explanation:

Tanya [424]3 years ago

3 0

B is the answer your welcome

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Write in normal form 2.8 x 10⁶ ?

Aliun [14]

It would be 2,800,000

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3 years ago

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If 2x+20 and 3x+20 makes a straight angke find value of x

erma4kov [3.2K]

Answer:

x = 28

Step-by-step explanation:

The sum of the angles in a straight angle = 180°

Sum the angles and equate to 180

2x + 20 + 3x + 20 = 180

5x + 40 = 180 ( subtract 40 from both sides )

5x = 140 ( divide both sides by 5 )

x = 28

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3 years ago

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach

nlexa [21]

Answer:

A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that $\mu = 130, \sigma = 30$

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 130}{30}$

$X - 130 = 0.675*30$

$X = 150.25$

A score of 150.25 is necessary to reach the 75th percentile.

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3 years ago

Which expression is equivalent to (1/16)^-4<br> Pls help 30 points timed

worty [1.4K]

The first one! It is using the distributive or associative property!

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3 years ago

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

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3 years ago