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3 years ago

Joshua ran 1.45 miles, and Jasmine ran 1.5 miles who Ran Farther?

Mathematics

2 answers:

Lilit [14]3 years ago

7 0

Jasmine ran faster because 1.45 is less than 1.50 which can also be written as 1.5

Natalija [7]3 years ago

6 0

Answer:

Jasmine

Step-by-step explanation:

You might be interested in

Four students are running for class president: liz, sam, sue, and tom. the probabilities of sam, sue, and tom winning are 7⁄25 ,

Natali5045456 [20]

First, we need to convert all the fractions to have the same denominator. 10 and 25 are both multiples of 100, so 100 would be appropriate.

Sam has a 7/25 chance. Because we want ?/100, something needs to change. To get from 25 to 100, you need to times 25 by 4, right? So, do the same with the 7.
7 x 4 = 28. Therefore Sam has a 28/100 chance.

Sue has 3/10. Using the same method, we can see that 3 needs to be multiplied by 10 (because 10 times 10 = 100). So Sue has a 30/100 chance.

Tom is already in the fraction we like, so just keep this as 21/100.

Now, add 28/100, 30/100 and 21/100 to get 79/100.

Because won of them will get the role of class president, we know that the probability adds to 1. To get a full probability (100/100, or 1), what needs to be added to 79/100?

Another way of going about this is 100/100-79/100. The answer is 21/100

The probability of Liz winning is 21/100.

Let me know if this is still unclear, I would be more than happy to explain in more detail if necessary :)

8 0

3 years ago

PLEASE HELP ME ON THIS, im begging you!

nexus9112 [7]

Answer:

question, what grade are you in? it might help me solve the problem

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve for the equation, show all your work. This is an absolute value equation. You need to show both equations and the work for

VMariaS [17]

Answer:

$x=-1$ or $7$

Step-by-step explanation:

We use casework on when $3x-9=12$ and when $3x-9=-12$ .

For the first case, $3x-9=12$ , we add 9 to both sides to get

$3x=21$ .

Dividing both sides by 3 gives

$x=7.$

For the second case, $3x-9=-12$ , we add 9 to both sides to get

$3x=-3$ .

Dividing both sides by 3 gives

$x=-1$ .

Checking both cases, we plug in $x=7$ and $x=-1$ .

For the first case, we have $|3(7)-9|=|21-9|=|12|=12$ , which satisfies the equation.

For the second case, we have $|3(-1)-9|=|-3-9|=|-12|=12$ , which also satisfies the equation.

This gives us two solutions to the equation; $x=7$ and $x=-1$ .

4 0

2 years ago

If the graph of the linear equations in a system are parallel what does that mean about possible solutions of the system?

professor190 [17]

Since, the system of linear equations are parallel

so, they will never cross or intersect each other

And we know that

solution can only be found if both lines intersect atleast one point

But here , lines are parallel

so, they never intersect each other throughout number line

Since, they never intersect

so, we won't get any common value between both lines

Hence,

There is no solution...........Answer

8 0

3 years ago

Each of six jars contains the same number of candies. Alice moves half of the candies from the first jar to the second jar. Then

tino4ka555 [31]

Answer:

The number of candies in the sixth jar is 42.

Step-by-step explanation:

Assume that there are <em>x</em> number of candies in each of the six jars.

⇒ After Alice moves half of the candies from the first jar to the second jar, the number of candies in the second jar is:

$\text{Number of candies in the 2nd jar}=x+\fracx}{2}=\frac{3}{2}x$

⇒ After Boris moves half of the candies from the second jar to the third jar, the number of candies in the third jar is:

$\text{Number of candies in the 3rd jar}=x+\frac{3x}{4}=\frac{7}{4}x$

⇒ After Clara moves half of the candies from the third jar to the fourth jar, the number of candies in the fourth jar is:

$\text{Number of candies in the 4th jar}=x+\frac{7x}{4}=\frac{15}{8}x$

⇒ After Dara moves half of the candies from the fourth jar to the fifth jar, the number of candies in the fifth jar is:

$\text{Number of candies in the 5th jar}=x+\frac{15x}{16}=\frac{31}{16}x$

⇒ After Ed moves half of the candies from the fifth jar to the sixth jar, the number of candies in the sixth jar is:

$\text{Number of candies in the 6th jar}=x+\frac{31x}{32}=\frac{63}{32}x$

Now, it is provided that at the end, 30 candies are in the fourth jar.

Compute the value of <em>x</em> as follows:

$\text{Number of candies in the 4th jar}=40\\\\\frac{15}{8}x=40\\\\x=\frac{40\times 8}{15}\\\\x=\frac{64}{3}$

Compute the number of candies in the sixth jar as follows:

$\text{Number of candies in the 6th jar}=\frac{63}{32}x\\$

$=\frac{63}{32}\times\frac{64}{3}\\\\=21\times2\\\\=42$

Thus, the number of candies in the sixth jar is 42.

4 0

3 years ago