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3 years ago

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The ratio of girls to boys in the junior high band is 5 to 7. At the beginning of year, there were 72 students in the band. By t

he end of the year, the ratio of girls to boys was 3 to 4. If there are now 48 boys in the band, how many girls joined the band during the school year?

2 answers:

Andre45 [30]3 years ago

8 0

Answer:

Your answer will be 36

Step-by-step explanation:

I know that because there were 30 girls in the beginning of school and 6 girls joined band with is 36

rusak2 [61]3 years ago

7 0

Ratio is 5girls:7boys with total 72 students
5+7= 12
72➗12= 6
6(5:7)= 30 girls : 42 boys

New ratio is 3:4 with 48 boys

48➗4= 12

So for girls 3x12= 36

So started with 30 girls ended with 36 girls... 6 girls joined band during the school year.

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Jenny 112 have a possible 148 points on a test she lost four points for each incorrect answer how many incurred answers does she

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she has nine incorrect answers

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3 years ago

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If f(x) =ax+b ,f(-3)=-4 and f(3)=2, find the value of a and b please help

hichkok12 [17]

Answer:

a = 1

b = -1

Step-by-step explanation:

First, let's make equations out of those given f(x)'s

-4 = -3a + b

2 = 3a + b

We can use elimination method to get rid of the a real quick

-4 = b

2 = b

Add those together

-2 = 2b

Divide both sides by 2

-2/2 = 2b/2

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Plug in the new b in one of the original equations; I'm gonna use the bottom one

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2 years ago

Goshia [24]

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3 years ago

Read 2 more answers

Help help I don’t really get this???

insens350 [35]

Answer:

Question 4: $y=\displaystyle -\frac{4}{5}x$

Question 5: $y=-5x-3$

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where <em>m</em> is the slope of the line and <em>b</em> is the y-intercept (the y-coordinate of the point where the line crosses the y-axis).

<u>Question 4</u>

<u>1) Determine the slope (</u><u><em>m</em></u><u>)</u>

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$ where two points that pass through the line are $(x_1,y_1)$ and $(x_2,y_2)$

In the graph, two easy-to-identify points on the line are (-5,4) and (5,-4). Plug these into the equation:

$\displaystyle m=\frac{-4-4}{5-(-5)}\\\\\displaystyle m=\frac{-4-4}{5+5}\\\\\displaystyle m=\frac{-8}{10}\\\\\displaystyle m=-\frac{4}{5}$

Therefore, the slope of the line is $\displaystyle -\frac{4}{5}$ . Plug this into $y=mx+b$ as the slope (<em>m</em>):

$y=\displaystyle -\frac{4}{5}x+b$

<u>2) Determine the y-intercept (</u><u><em>b</em></u><u>)</u>

On the graph, we can see that the line crosses the y-axis when y is 0. Therefore, the y-intercept (<em>b</em>) is 0. Plug this into $y=\displaystyle -\frac{4}{5}x+b$ :

$y=\displaystyle -\frac{4}{5}x+0\\\\y=\displaystyle -\frac{4}{5}x$

<u>Question 5</u>

<u>1) Determine the slope (</u><u><em>m</em></u><u>)</u>

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Two easy-to-identify points are (-1,2) and (0,-3). Plug these into the equation:

$\displaystyle m=\frac{-3-2}{0-(-1)}\\\\\displaystyle m=\frac{-3-2}{0+1}\\\\\displaystyle m=\frac{-5}{1}\\\\m=-5$

Therefore, the slope is -5. Plug this into $y=mx+b$ :

$y=-5x+b$

<u>2) Determine the y-intercept (</u><u><em>b</em></u><u>)</u>

On the graph, we can see that the line crosses the y-axis at the point (0,-3). The y-coordinate of this point is -3. Therefore, the y-intercept (<em>b</em>) is -3. Plug this into $y=-5x+b$ :

$y=-5x+(-3)\\y=-5x-3$

I hope this helps!

8 0

2 years ago

Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a

krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. <u>Finding z-score for Reagan;</u>

Let X = distribution of SAT scores

SO, X ~ Normal( $\mu=1000, \sigma^{2}=100^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1000

$\sigma$ = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = $\frac{1140-1000}{100}$ = 1.4

2. <u>Finding z-score for Jessie;</u>

Let X = distribution of ACT scores

SO, X ~ Normal( $\mu=17, \sigma^{2}=5^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 17

$\sigma$ = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = $\frac{30-17}{5}$ = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

6 0

3 years ago