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3 years ago

6

Your current test average in Biology is a 73. If the standard deviation is 25.1, you are a consistent Biology student.

1 answer:

Ugo [173]3 years ago

5 0

Your answer is true! I hope that helps ;)

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Please help me i have to finnish this!!!

sleet_krkn [62]

Answer:

488

Step-by-step explanation:

$6x^{2} + 2$ = 0
replace x = -9 => 6 x (-9)^2 + 2 = 488

8 0

3 years ago

Simplify 3/8 squared

8_murik_8 [283]

Answer:

9/64

Step-by-step explanation:

3 x 3 is 9 and 8 x 8 is 64 and this fraction cannot be simplified

3 0

3 years ago

Read 2 more answers

This is due like right now, please help me!!!!

Varvara68 [4.7K]

$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

<u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
<u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
<u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

$\bold{\underline{ Let's \: Begin}}$

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

$\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}$

$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

$\sf{ c = 60 \: and \: e = 51 }$

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }$

$\sf{ 158 + b = 156 + d = 166 + a = 189}$

<u>Now</u><u>, </u>

$\sf{ For \: b, 158 + b = 189 }$

$\sf{ b = 189 - 158 }$

$\sf{ b = 31}$

$\sf{ For \: d , 156 + b = 189 }$

$\sf{ d = 189 - 156 }$

$\sf{ d = 33}$

$\sf{ For \: a, 166 + a = 189 }$

$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

8 0

3 years ago

Factor by Grouping.

Komok [63]

Answer:

The ac product of $35x^2+41x+12$ is <u>420</u>.

The factors of the ac product that add to 41 are <u>20</u> and <u>21</u>.

$35x^2+41x+12 =$ $(7x+4)(5x+3)$

Step-by-step explanation:

1) The general form of a quadratic is $ax^2 + bx + c$ . Hence, multiplying 35 by 12 gives you the product of ac, which is 420.

2) We need to find two numbers that multiply to 420 and add up to 41 <u>simultaneously</u>. If we pull out the factors of 420, two of them will be 20 and 21, which multiply to 420 as well as add up to 41.

3) Write $41x$ as a sum.

$35x^2+20x+21x+12$

Now, we can factor them out by grouping.

$35x^2+20x+21x+12\\5x(7x+4)+3(7x+4)$

Since, $7x +4$ is common in both of the factors, we only take one of the $7x+4$ along with $5x$ and $3$ .

$(7x+4)(5x+3)$

6 0

2 years ago

Read 2 more answers

What is the length of a hypotenuse of the right triangle shown

JulijaS [17]

Answer:

No answer (NO PICTURE)

Step-by-step explanation:

5 0

3 years ago