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3 years ago

Math help please? <3

Mathematics

2 answers:

Alex73 [517]3 years ago

8 0

The first one :) x>3 x<1

asambeis [7]3 years ago

3 0

A has no answer as you see in the pic

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What is two thousand, six hundred nine dollars and seventy five cents written in expanded notation

Gnesinka [82]

Answer:

2000+600+9+.75 maybe?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Solve the following system of equations algebraically 3x-y=0 5x+2y=22

Lyrx [107]

3x - y = 0 ⇒ 2(3x - y = 0 ) ⇒ 6x - 2y = 0

5x + 2y = 22 ⇒ 1(5x + 2y = 22) ⇒ 5x + 2y = 22

11x = 22

x = 2

3x - y = 0 ⇒ 3(2) - y = 0 ⇒ 6 - y = 0 ⇒ 6 = y

Answer: x=2, y=6

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4 years ago

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

3 0

4 years ago

Gwen measured a kitchen and made a scale drawing. The scale she used was 11 inches : 5 feet. If a countertop in the kitchen is 2

aliina [53]

Answer:

11x 5 =55-22=33

Step by step explanation

i think it is useful nark me as branlist

8 0

3 years ago

The number of motor vehicles registered in millions in the US has grown as follows:

zysi [14]

Answer:

a) Figure attached

b) $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=10 $\sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274$

$r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.819)^2][10(192.8274) -(43.5231)^2]}}=0.989$

So then the correlation coefficient would be r =0.989

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

Solution to the problem

Part a

Year (x): 1940, 1945 1950, 1955, 1960, 1965, 1970, 1975, 1980, 1985

Vehicles (Y): 32.4, 31.0, 49.2, 62.7, 73.9, 90.4, 108.4, 132.9, 155.8, 171.7

After apply natural log for the two variables and create the scatterplot in excel we got the following result on the figure attached.

Part b

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=10 $\sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274$

$r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.81948)^2][10(192.8274) -(43.5231)^2]}}=0.989$

So then the correlation coefficient would be r =0.989

4 0

3 years ago