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irga5000 [103]
3 years ago
9

Use intercepts to graph the equation 2x-5y=10 Show work :)

Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0
Ok so change the questions to -5y=10-2x then divide both side 5y=-10+2x then record the terms y=-2+2/5x the answer is y=2/5x-2
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A fish swam 75 feet to the bottom of the river. The fish then swam up 35 feet towards the surface of the river and stopped. How
telo118 [61]

Answer: 40

Step-by-step explanation:

75-35

5 0
3 years ago
Question 2 (1 point)
jonny [76]

Answer:

(2,4)

Step-by-step explanation:

Equation 1:

x+y=6(for the 6 times they scored)

x=-y+6

Equation 2:

7x+3y=26 (for the 7 point touchdowns, and 3 point field goals.)

7(-y+6)+3y=26(replace x with -y+6)

7y+42+3y=26(distrubite property)

-4y+42=26

-4y= -16

y= 4

Now that we know that y is equal to 4:

x+4=6

x=2

5 0
3 years ago
Read 2 more answers
PLEASE HELP with these math questions (Please don't answer if you don't know all of them)
Leona [35]
Sqrt (53) = 10 * sqrt (0.53)

0.53 = 64/121
sqrt (0.53) = sqrt (64)/ sqrt (121) = 8/11 = 0.7273
Therefore sqrt (53) = 10 * 0.7273 = 7.27

sqrt (108) = 10 * sqrt (1.08)
sqrt (1.08) = sqrt (676/625) = 26/25 = 1.04
Therefore sqrt (108) = 10 * 1.04 = 10.4

sqrt (128) = 10 * sqrt (1.28)
sqrt (1.28) = sqrt (289/225) = 17/15 = 1.133
Therefore sqrt (108) = 10 * 1.133 = 11.33

5 0
3 years ago
The area of the square, the
const2013 [10]

Answer:

B.

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.  

Please have a look at the attached photo.  

My answer:

As given in the question, we know that:

The ratio of the area of the circle to the area  of the square is π/4

  • The formula to find the volume of the cone is:

V = 1/3*the height*the base area

<=> V1 = 1/3*h*πr^{2}

  • The formula to find the volume of the pyramid is:

V2 = 1/3*the height*the base area

<=> V =  1/3*h*4r^{2}

=> the ratio of volume of the cone to the pyramid is:

= \frac{V1}{V2}

= (1/3*h*πr^{2} ) / ( 1/3*h*4r^{2} )

= π/4

S we can conclude that the volume of the cone equals π/4  the volume of the  pyramid

Hope it will find you well.

7 0
3 years ago
Examine the required sample size needed to be able to
Xelga [282]

Answer:

n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

ME=2.33\frac{500}{\sqrt{2000}}=26.05  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=500 represent the population standard deviation assumed

n represent the sample size  (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got z_{\alpha/2}=2.33, replacing into formula (b) we got:

n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

ME=2.33\frac{500}{\sqrt{2000}}=26.05    

3 0
4 years ago
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