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Diano4ka-milaya [45]

3 years ago

6

What is the slope of the line given by the equation -3x = 30 - 5y?

1 answer:

zaharov [31]3 years ago

8 0

Answer:

The slope of the line given by the equation -3x=30-5y is $\frac{3}{5}$ .

Step-by-step explanation:

-3x=30-5y

Move the -3x to the right side of the equal sign and the -5y to the left.

5y=30+3x

By switching them to opposite sides, their sign changes from - to +.

Then, divide both sides by 5 in order to isolate the y.

$\frac{5y}{5} = \frac{30+3x}{5}$

You get:

<em>y= </em> $\frac{3}{5} x +6$ <em />

According to y=mx+b, mx refers to the slope.

In this case, the slope would be $\frac{3}{5}$ .

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Last year Lira earned $12,000 less than her husband Todd. Together they earned $75,000. How much did Lira earn last year?

nexus9112 [7]

Given:

Last year Lira earned $12,000 less than her husband Todd.

Together they earned $75,000.

To find:

The amount earned by Lira in last year.

Solution:

Let x be the amount earned by Lira's husband inn last year.

Last year Lira earned $12,000 less than her husband Todd.

Amount earned by Lira = $x-12000$

Total amount earned by Lira and her husband = $x+(x-12000)$

= $2x-12000$

Together they earned $75,000.

$2x-12000=75000$

$2x=75000+12000$

$2x=87000$

$x=\dfrac{87000}{2}$

$x=43500$

Now,

Amount earned by Lira = $x-12000$

= $43500-12000$

= $31500$

Therefore, the amount earned by Lira in last year is $31500.

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If a common logarithm is written without a base, what base should be used in evaluating it?

Mice21 [21]

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3 years ago

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6 0

3 years ago

The probability distribution of the amount of memory X (GB) in a purchased flash drive is given below. x 1 2 4 8 16 p(x) .05 .10

Rama09 [41]

Answer:

a) $E(X) = 6.45$

b) $E(X^{2} )= 57.25$

c) $V(X) = 15.648$

d) E(3X + 2) = 21.35

e) $E(3X^{2} +2) = 173.75$

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) $E(X) = \sum xP(x)$

$E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45$

b)

$E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\ E(X^{2} )= 57.25$

c)

$V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648$

d)

$E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35$

e)

$E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75$

f)

$V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832$

g)

$E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45$

h)

$V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648$

6 0

3 years ago