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3 years ago

Stephanie gets on her bike and rides for 3 hours. She rides at an average speed of 12.5 km per

1 answer:

8 0

Stephanie rode the bike with an speed of 37.5 km. If she ride the bude and the average speed for her is 12.5 km per hour, multiply 12.5 with how much hour to get the answer. 12.5x3=37.5. This means that Stephanime rode her bike and got a speed of 37.5 km.

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Guys, help me out, please.

goldfiish [28.3K]

Answer:

Step-by-step explanation:

Given

$\frac{28}{3+4}$ ← evaluate the denominator

= $\frac{28}{7}$ ← perform the division

= 4

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3 years ago

The 4 - pack is priced at £2.04

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The 4 Pack is the better value ;)

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3 years ago

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Leni [432]

Answer:

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3 years ago

Read 2 more answers

Jim worked 40 hours last week, plus 8 overtime hours at the time-and-a-half rate. His gross pay was $884. A. What was his hourly

marishachu [46]

Answer:

Hourly rate: $17.

Hourly overtime rate: $25.5.

Step-by-step explanation:

Let x represent Jim's hourly pay.

We have been given that Jim worked 40 hours last week. So amount paid in 40 hours would be $40x$ .

We are also told that his over time pay is time-and-a-half rate that is $1.5x$ .

The amount earned for working 8 overtime hours would be $8\cdot 1.5x=12x$ .

Since Jim's gross pay was $884, so we will equate amount earned in 40 hours and 8 hours of overtime with 884 as:

$40x+12x=884$

$52x=884$

$\frac{52x}{52}=\frac{884}{52}$

$x=17$

Therefore, Jim's hourly rate is $17.

Overtime rate would be $1.5x\Rightarrow1.5(17)=25.5$

Therefore, Jim's hourly overtime rate is $25.5.

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3 years ago

the set $s = \$ contains the first $50$ positive integers. after the multiples of 2 and the multiples of 3 are removed, how many

Sophie [7]

Clearly, |S| = 50.

Count the multiples of 2 between 1 and 50:

⌊50/2⌋ = ⌊25⌋ = 25

(where ⌊x⌋ denotes the "floor of x", or the largest integer that is smaller than or equal to x; in other words, round <u>down</u> to the nearest integer)

Count the multiples of 3 between 1 and 50:

⌊50/3⌋ ≈ ⌊16.667⌋ = 16

Since LCM(2, 3) = 6, the sets of multiples of 2 and multiples of 3 have some overlap. Count the multiples of 6 between 1 and 50:

⌊50/6⌋ ≈ ⌊8.333⌋ = 8

Then by the inclusion/exclusion principle, we remove from S

25 + 16 - 8 = 33

elements, so that the new set S contains 50 - 33 = 17 elements.

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3 years ago