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3 years ago

One lap around the school track is 5/8 of a mile Carin ran 3 1/2 laps how far did she run

Mathematics

1 answer:

bearhunter [10]3 years ago

4 0

Given length of a lap = 5/8 mile.

Number of laps Carin ran = 3 1/2 laps.

3 1/2 can be read as 3 and a half. In improper fractions, it can be written as

(3*2+1)/2 = 7/2 laps.

In order to find the total distance covered in 3 1/2 or 7/2 laps, we need to multiply length of a lap by total number of laps ran.

Therefore, total distance covered = 7/2 × 5/8.

= $\frac{7}{2}\times\frac{5}{8}$

Multiplying across, we get

= $\frac{7\times5}{2\times8}$

= $\frac{35}{16}$

Let us convert 35/16 into mixed fractions.

On dividing 35 by 16, we get 2 quotient and 3 remainder.

So, the mixed fraction would be 2 3/16.

Therefore, she ran 2 3/16 miles in 3 1/2 laps.

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I don't know how to do this or what i'm doing plz help

mote1985 [20]

recalling that d = rt, distance = rate * time.

we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}$

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}$

$\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}$

7 0

3 years ago

How to calculate rational numbers

Anuta_ua [19.1K]

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Hi my lil bunny!

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The simplest method to find a rational number between two rational numbers x and y is to divide their sum by 2.

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Have a great day/night!

❀*May*❀

8 0

3 years ago

Which phrase represents' the sum of 11 and a number

denis23 [38]

Answer:

11 + n

Step-by-step explanation:

sum means addition, so u need a + sign

n is usually the variable used for 'a number'

so 11 and (+) a number (n) = 11+ n

hope this helps!

5 0

3 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago

A test is used to assess readiness for college. In a recent year, the mean test score was 20.3 and the standard deviation was 4

Vlada [557]

Answer:

$\mu = 20.3$

$\sigma = 4.9$

And we can find the limits in order to consider values as significantly low and high like this:

$Low\leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5$

$High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can consider a value to be significantly low if we have that the z score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is higher tor equal to 2.

For this case we have the mean and the deviation given:

$\mu = 20.3$

$\sigma = 4.9$

And we can find the limits in order to consider values as significantly low and high like this:

$Low \leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5$

$High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1$

6 0

3 years ago