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3 years ago

15

Mrs. Castillo's class has 15 boys and 5 girls. 40% of the class rides the bus to school. How many students ride the bus to schoo

l?
Group of answer choices

8

12

60

6

1 answer:

Anna35 [415]3 years ago

3 0

I think it’s 8 I’m not sure

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Fill in the blank:<br><br><br>5/12 * __ = 1

Kisachek [45]

Answer:

2.4

Step-by-step explanation:

5/12 * _ = 1

0.41666666666 x 2.4 = 1

5 0

3 years ago

Given the information below, write the equation in factored form.

ser-zykov [4K]

Answer

y = 2(x - 8)(x + 2)

Step-by-step explanation:

A(x - 8)(x + 2)

A × -8 × 2 = -32

-16A = -32

A = 2

y = 2(x - 8)(x + 2)

4 0

3 years ago

What are the minimum, first quartile, median, third quartile, and maximum of the data set? 2, 13, 17, 14, 9, 3, 16, 12

Alex777 [14]

So before anything, rearrange the data so that it's in ascending order: {2,3,9,12,13,14,16,17}

Next, the minimum and maximum are as they seem: the smallest and largest numbers in the data. Looking at our data, <u>2 is our minimum and 17 is our maximum.</u>

Next, to find the median, find the number that is in the middle of the data set. If there isn't one number in the middle, find the average of the two middle numbers to get your median:

$\{2,3,9,\overbrace{\boxed{12,13,}}^{\textsf{middle numbers}}14,16,17\}\\\\\frac{12+13}{2}\\\\\frac{25}{2}\\\\12.5$

<u>The median is 12.5.</u>

Next, to find the first quartile, or the lower quartile, find the "median" of the numbers to the left of the median.

$\overbrace{\{2,\boxed{3,9,}12,}^{\textsf{left of median}}13,14,16,17\}\\\\\frac{3+9}{2}\\\\\frac{12}{2}\\\\6$

<u>The first quartile is 6.</u>

Next, to find the third quartile, or the upper quartile, find the "median" of the numbers to the right of the median.

$\{2,3,9,12,\overbrace{13,\boxed{14,16,}17\}}^{\textsf{right of median}}\\\\\frac{14+16}{2}\\\\\frac{30}{2}\\\\15$

<u>The third quartile is 15.</u>

3 0

3 years ago

Please help me on this

alexandr402 [8]

Answer:

5/6 so the missing number is 5.

Step-by-step explanation: I hope this helps. :)

6 0

3 years ago

Read 2 more answers

Help I’m not sure with this one!! Urgent help

Nastasia [14]

This system has exactly one solution, because the two equation are linearly independent (i.e. one is not a multiple of the other).

We can explicitly find the solutions: from the first equation we deduce

$x = \dfrac{c-by}{a}$

Plug this value for x in the second equation:

$\dfrac{a}{3}\cdot \dfrac{c-by}{a}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{c-by}{3}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{2(c-by)+3by}{6}=\dfrac{c}{6}$

Multiply both sides by 6 to get

$2(c-by)+3by=c \iff 2c-2by+3by=c \iff by=-c \iff y=-\dfrac{c}{b}$

Plug this value for y in the expression for x:

$x = \dfrac{c-by}{a} = \dfrac{c-b\cdot\frac{-c}{b}}{a}=\dfrac{2c}{a}$

Since both a,b are not zero, both solutions are well defined.

8 0

3 years ago