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3 years ago

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» Quinn is a contestant on a quiz show. Quinn's score changes by -15 points for 4 questions in a row. D *) What is the total cha

nge in Quinn's score?

1 answer:

notka56 [123]3 years ago

6 0

Answer:

Quinn is a contestant on a quiz show. Quinn's score changes by -15 points for 4 questions in a row. D *) What is the total change in Quinn's score?

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I NEED HELP! Write an equation in standard form of the line

nadezda [96]

Answers:

First box = 1
Second box = -5
Third box = -15

This forms the equation 1x-5y = -15, which is the same as x-5y = -15

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Explanation:

Let's find the slope. I'll pick the first two rows to form the two points needed

(x1,y1) = (0,3)

(x2,y2) = (10,5)

These are then plugged into the slope formula.

m = slope

m = (y2-y1)/(x2-x1)

m = (5-3)/(10-0)

m = 2/10

m = 1/5

m = 0.2

The y intercept is b = 3 since this is the y value when x = 0.

With m = 0.2 and b = 3, we go from y = mx+b to y = 0.2x+3

Let's multiply both sides by 5 to turn that decimal value into a whole number. I'm picking 5 since 0.2 = 1/5

So we get

y = 0.2x+3

5y = 5*(0.2x+3)

5y = 5*0.2x+5*3

5y = x+15

Now let's get x and y together, but move that 15 to its own side

5y = x+15

x+15 = 5y

x = 5y-15

x-5y = -15

This is one way to write the equation in standard form.

Standard form is Ax+By = C, where A,B,C are integers. Some math textbooks insist that A > 0. In this case, A = 1, B = -5, and C = -15.

----------------------

As a way to check, we can plug in (x,y) = (15,6) from the third row of the table to see that..

x-5y = -15

15-5(6) = -15

15 - 30 = -15

-15 = -15

This verifies the third row. I'll let you check the first two rows of the table to fully confirm this equation works.

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2 years ago

How can you use ratios and rates to solve real- world problems?

larisa [96]

Usually, They make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane.

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3 years ago

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

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3 years ago

Domain and range???!!!!!!

elena55 [62]

Answer:

domain={1,-3}

range={-5,1,2,4}

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3 years ago

What is 23*536 because I don't know how and 54*653

Lubov Fominskaja [6]

23.536 54.653 it is decimal form

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2 years ago