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2 years ago

14

PLEASE HELP ME! WILL MARK BRAINlIEST!!

1 answer:

vivado [14]2 years ago

6 0

Answer:

1.) Increase

2.) Decrease

3.) Increase

4.) Decrease

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Alexa worked 30 hours last week and earned $450. Her pay has now been increased by 20%. Which equation describes her new rate of

vesna_86 [32]

Answer:

$p=18h$

or

$p=\dfrac{450}{30}\cdot 1.2\cdot h$

Step-by-step explanation:

If Alexa worked 30 hours last week and earned $450, then her hourly payment is

$\dfrac{\$450}{30}=\$15.$

Her payment has now been increased by 20%, so her hourly payment becomes

$\$15\cdot 1.2=\$18.$

If she worked h hours, then she earned $\$18h.$ If her new payment is p, then

$p=18h$

or

$p=\dfrac{450}{30}\cdot 1.2\cdot h.$

4 0

2 years ago

The joint probability mass function of XX and YY is given by p(1,1)=0p(2,1)=0.1p(3,1)=0.05p(1,2)=0.05p(2,2)=0.3p(3,2)=0.1p(1,3)=

inessss [21]

Answer:

P(Y=1|X=3)=0.125

Step-by-step explanation:

Given :

p(1,1)=0

p(2,1)=0.1

p(3,1)=0.05

p(1,2)=0.05

p(2,2)=0.3

p(3,2)=0.1

p(1,3)=0.05

p(2,3)=0.1

p(3,3)=0.25

Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)

P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)

P(X=3)=p(3,1) +p(3,2) +p(3,3)

P(X=3)=0.05+0.1+0.25=0.4

$P(Y=1|X=3)=\frac{P(X=3,Y=1)}{P(X=3)} =\frac{p(3,1)}{P(X=3)}=\frac{0.05}{0.4}= 0.125$

Hence P(Y=1|X=3)=0.125

8 0

2 years ago

3 of 5<br> What is the nth term rule of the linear sequence below?<br> 13,7,1, -5, -11, ...<br> © T=

OverLord2011 [107]

Answer:

The nth term is 6 .

13 — 7 = 6

7— 4 = 6

1 — (–5) = 6

-5— (-11) = 6

3 0

1 year ago

1468:22-6+8888-998=????????

irinina [24]

Answer is 7950,72. You can solve it's task on calculator.

7 0

2 years ago

Plz write this on paper help me and send it❤️

vovangra [49]

Answer:

1. $27^{\frac{2}{3} } =9$

2. $\sqrt{36^{3} } =216$

3. $(-243)^{\frac{3}{5} } =-27$

4. $40^{\frac{2}{3}}=4\sqrt[3]{25} =4325$

5. Step 4: $(\frac{343}{27}) ^{-1} =\frac{27}{343}$

6. $D. -72cd^{7}$

Step-by-step explanation:

Use the following properties:

$a^{\frac{x}{y} } =\sqrt[x]{a^{y} }$

$\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}$

$a^{-n} =\frac{1}{a^{n} }$

$(xy)^{z} =x^{z} y^{z} \\\\$

$(x^{y}) ^{z} =x^{yz}$

$x^{y} x^{z} =x^{y+z}$

So:

1. $27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9$

2. $\sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36} \sqrt{36} =6*6*6=216$

3. $(-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27$

4. $40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325$

5. $(\frac{343}{27}) ^{-1} =\frac{1}{\frac{343}{27} } =\frac{27}{343}$

6.

$(-8c^{9} d^{-3} )^{\frac{1}{3} } *(6c^{-1}d^{4})^{2} =\sqrt[3]{-8} c^{3} d^{-1} 36c^{-2} d^{8} \\\\-2c^{3} d^{-1} 36c^{-2} d^{8}=-72cd^{7}$

7 0

2 years ago