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3 years ago

Mia drove 824 miles on 34.8 gallons of gas. What was her gas mileage, rounded to the nearest mpg?

Mathematics

2 answers:

Leto [7]3 years ago

3 0

The correct answer is 24 mpg (rounded).
Hope this helps.

rusak2 [61]3 years ago

3 0

Her mpg rounded to the nearest thenth is 23.67 OR 23.7 MPG
hOPE THIS HELPS

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What does 10-|-5|=? please answer that would be appreciated

erik [133]

5 is the awser I'm pretty sure

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3 years ago

10 Students were surveyed.

katrin2010 [14]

Answer:mode

Step-by-step explanation:5+4+9+9+3+75+1=mode

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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.

5 0

3 years ago

The unit digit of square of the square of the number 42 is <br>

cupoosta [38]

Um so 42 squared is 1764 and 1764 squared is 3111696 so then 6 is the unit digit

7 0

3 years ago

When y=-2, what kind of line does it have horizontal or vertical

Rashid [163]

Answer: Horizontal

Step-by-step explanation: The equation <em>y = -2</em> can be thought of as y = 0x - 2. So our line has a slope of 0 and a y-intercept of -2.

To graph it, we start with the y-intercept, down 2 units on the y-axis. Now, if the slope of a line is 0, then the line must be flat or horizontal.

So we just draw a horizontal line through the y-intercept of -2.

In fact, when the equation of any line reads y = a number, it's graph will always be a horizontal line. For example, y = 3, y = -10, y = -8 and so on.

Image provided below.

8 0

3 years ago