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4 years ago

HELLLLLLLLLLLLLLLLLLLPPPPPPPPPP 99PTS

Mathematics

1 answer:

Elena-2011 [213]4 years ago

6 0

A graphing calculator can be helpful for these.

A) Since neither inequality includes the "or equal to" case, the boundary line for the solution set is a dashed line. For the first inequality, the shading is above the line; for the second, the shading is below the line. The solution set is points to the right of perpendicular lines with slope -2 and +1/2 through point (2, -1).

B) The point -2, 3 satisfies, neither inequality.
.. 3 > -2(-2) +3 . . . FALSE
.. 3 < 1/2(-2) -2 . . .FALSE

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Can 6/-6 be simplified?

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Answer:

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Step-by-step explanation:

Because fractions are technically division, 6 ÷-6 = -1

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(1 point) Use Euler's method with step size 0.25 to compute the approximate y-values y1, y2, y3, and y4 of the solution of the i

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Answer:

$y_1=4.75\\y_2=10.625\\y_3=22.5\\y_4=46.375$

Step-by-step explanation:

Formula to calculate yn using Euler method is as follows:

$x_{n+1}=x_n+h\\\\h=step\,\,size\\\\y_{n+1}=y_n+F(x_n,y_n)\\\\where\\\\y'=F(x,y)\,\, and\,\,y(x_o)=y_o$

Given that:

$y'=2+2x+4y\\\\x_o=1\,,\,y_o=2\,,\,h=0.25$

$F(x_o,y_o)=1+2(1)+4(2) =11\\\\y_1=y_o+hF(x_o,y_o)\\\\y_1=2+(0.25)(11)=4.75\\\\x_1=1+0.25=1.25\\\\F(x_1,y_1)=2+2(1.25)+4(4.75)= 23.5\\\\y_2=y_1+hF(x_1,y_1)\\\\y_2=4.75+(0.25)(23.5)=10.625\\\\x_2=1.25+.25=1.5\\\\f(x_2,y_2)=2+2(1.5)+4(10.625)=47.5\\\\y_3=y_2+h(F(x_2,y_2)\\\\y_3=10.625+(0.25)(47.5)=22.5\\\\x_3=1.5+0.25=1.75\\\\F(x_3,y_3)=2+2(1.75)+4(22.5)=95.5\\\\y_4=y_3+hF(x_3,y_3)\\\\y_4=22.5+0.25(95.5)=46.375$

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$10.20 for each CD before tax.
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3 years ago

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Step-by-step explanation:

Problems of normally distributed samples are 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:

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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:

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