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Feliz [49]
3 years ago
15

A running back is having a great day on the football field. It’s only the third quarter, and he’s already rushed for 203 yards.

He’s close to breaking his previous record of 229 rushing yards in a game. How many yards does he need to gain during the rest of the game to break his record?
Mathematics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

He needs to gain more than 26 yards in order to break his record

Step-by-step explanation:

Let y stand for the number of yards he’ll have to gain. The sum of y plus the number of yards he already has (203) must be greater than 229 in order for him to break the record.  

203 + y > 229                           Set up the inequality.

203 − 203 + y > 229 − 203       Subtract 203 from both sides of the inequality.

y > 26

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navik [9.2K]

Answer:

Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Step-by-step explanation:

<u>Step 1:  Multiply</u>

<u />(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\

(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}

(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x}  )

2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}

10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Answer:  Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

6 0
4 years ago
Read 2 more answers
Jessica, Lauren and Kimmie were paid a total of $120 for babysitting in a ratio of 2:3:5. How much less did Jessica make than Ki
Tju [1.3M]

Answer:

Jessica makes $36 less than Kimmie  

Step-by-step explanation:

Let

x -----> amount paid to Jessica

y  ----> amount paid to Lauren

z ----> amount paid to Kimmie

we know that

x+y+z=120 ------> equation A

\frac{x}{y}=\frac{2}{3}

x=\frac{2}{3}y -----> equation B

\frac{y}{z}=\frac{3}{5}

z=\frac{5}{3}y -----> equation C

substitute equation B and equation C in equation A and solve for y

\frac{2}{3}y+y+\frac{5}{3}y=120

\frac{10}{3}y=120

y=120(3)/10

y=36

<em>Find the value of x</em>

x=\frac{2}{3}(36)=24

<em>Find the value of z</em>

z=\frac{5}{3}(36)=60

therefore

The  amount paid to Jessica was $24

The amount paid to Lauren was $36

The amount paid to Kimmie was $60

To find out how much less Jessica did than Kimmie, subtract the amount paid to Jessica from the amount paid to Kimmie.

\$60-\$24=\$36

therefore

Jessica makes $36 less than Kimmie

7 0
3 years ago
Newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean
Maksim231197 [3]

The correct option is:   a female who weighs 1500 g

<em><u>Explanation</u></em>

<u>Formula for finding the z-score</u> is:  z= \frac{X-\mu}{\sigma}

Newborn males have weights with a mean(\mu) of 3272.8 g and a standard deviation(\sigma) of 660.2 g.

So, the z-score for the newborn male who weighs 1500 g will be.......

z(X=1500)=\frac{1500-3272.8}{660.2}=-2.685... \approx -2.69

According to the normal distribution table,  P(z=-2.69)=0.0036 = 0.36\%

Now, newborn females have weights with a mean(\mu) of 3037.1 g and a standard deviation(\sigma) of 706.3 g.


So, the z-score for the newborn female who weighs 1500 g will be.......

z(X=1500)=\frac{1500-3037.1}{706.3}=-2.176... \approx -2.18

According to the normal distribution table,  P(z=-2.18)=0.0146 = 1.46\%

As we can see that the <u>probability that a newborn female has weight of 1500 g is greater than newborn male</u>,  so a newborn female has the weight of 1500 g that is more extreme relative to the group from which he came.

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Step-by-step explanation:

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Answer:

See the proof below

Step-by-step explanation:

Let the line AB be a straight line on the parallelogram.

A dissection of the line (using the perpendicular line X) gives:

AY ≅ BX

Another way will be using the angles.

The angles are equal - vertically opposite angles

Hence the line  AY ≅ BX (Proved)

8 0
3 years ago
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