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3 years ago

Can someone please help me?

Mathematics

1 answer:

Levart [38]3 years ago

6 0

a) residential plan cost more by $4

b) it costs the same at 14 HCF. If they use less than this, then residential plan costs less

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The scale of a map is 6 km = 4 mi. What is the length on the map if the actual is 40.1<br> miles.

beks73 [17]

Answer:

60.15 km

Step-by-step explanation:

If 6 km on the map represents 4 mi on land, then we'd get the answer by using the relation

6 km -> 4 mi

x km -> 40.1 mi

Where x km is the length on the map we're looking for. If we cross multiply

6 * 40.1 = 4x

240.6 = 4x

x = 240.6/4

x = 60.15 km.

That is the length we're looking for

8 0

2 years ago

There were 90 runners just start a race in the first half of the race 1/3 of them dropped out in the second half of the race 1/5

Eva8 [605]

Answer:

6 runners finished the race

Step-by-step explanation:

it starts with 90 runners. then 1/3 of them dropped out in the second half of the race, that means you do 90 divided by 3

90 ÷ 3 = 30

Then, 1/5 of the runners remained.

That means you do 30 divided by 5

30÷ 5= 6

so the answer is 6 remaining runners finished the race.

8 0

2 years ago

Which of the following statements are true? Select all that apply.

Tom [10]

Answer: The statements in options 2 and 4 are true.

Explanation:

The LCM of two numbers is the smallest number that is multiple of both numbers.

The factor of 8 and 12 are,

$8=2\times 2\times 2$

$12=2\times 2\times 3$

Since 2 and 2 are common in both but 2 and 3 are not same, therefore the L.C.M. of 8 and 12 is,

$L.C.M.(8,12)=2\times 2\times 2\times 3=24$

Therefore the statement in option 1 is incorrect.

The factor of 6 and 9 are,

$6=2\times 3$

$9=3\times 3$

Since 3 is common in both but 2 and 3 are not same, therefore the L.C.M. of 6 and 9 is,

$L.C.M.(6,9)=2\times 3\times 3=18$

Therefore the statement in option 2 is correct.

The factor of 11 and 4 are,

$11=1\times 11$

$4=2\times 2$

Since all factor are different, therefore the L.C.M. of 11 and 4 is,

$L.C.M.(11,4)=1\times 11\times 2\times 2=44$

Therefore the statement in option 3 is incorrect.

The factor of 9 and 10 are,

$9=1\times 9$

$10=2\times 5$

Since all factor are different, therefore the L.C.M. of 9 and 10 is,

$L.C.M.(9,10)=1\times 9\times 2\times 5=90$

Therefore the statement in option 4 is correct.

4 0

2 years ago

A total of 1 232 students have taken a course in Spanish, 879 have taken a course in French, and 114 have taken a course in Russ

ElenaW [278]

Answer:

$n(S\cap F \cap R)=7$

Step-by-step explanation:

The Universal Set, n(U)=2092

$n(S)=1232\\n(F)=879\\n(R)=114$

$n(S\cap R)=23\\n(S\cap F)=103\\n(F\cap R)=14$

Let the number who take all three subjects, $n(S\cap F \cap R)=x$

Note that in the Venn Diagram, we have subtracted $n(S\cap F \cap R)=x$ from each of the intersection of two sets.

The next step is to determine the number of students who study only each of the courses.

$n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x$

These values are substituted in the second Venn diagram

Adding up all the values

2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]

2092=2085+x

x=2092-2085

x=7

The number of students who have taken courses in all three subjects, $n(S\cap F \cap R)=7$

3 0

3 years ago

What fraction is bigger 14/7 or 2/4

Arisa [49]

14/7, 14/7=2 and 2/4= 0.5

7 0

2 years ago

Read 2 more answers