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lukranit [14]
4 years ago
10

H(0)=

Mathematics
1 answer:
Goshia [24]4 years ago
6 0

Answer:

I feel like it b hope it correct though

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A family discount card offers deals for the cinema. The card states:
Rufina [12.5K]

Answer:

a)£21.80

b) 3 trips

Step-by-step explanation:

The cost of an adult is $10.

A child’s ticket is $6.

FAMILY DISCOUNT CARD

- 45% off all adults tickets

- Children save 2/5 off of full price tickets

a) How much will it cost to buy 2 adult tickets and children’s tickets?

(a)

Adults: 2 × £10 = £20

(45 ÷ 100) x £20 = £9

£20 - £9 = £11

Children: 3 × £6 = £18

2 /5 of £18 = £7.20

£18 - £7.20 = £10.80

Total: £11 + £10.80 = £21.80

b) The cost of a family discount card is $40 per year. How many times in the same year would 2 adults and 3 children need to go before you start to savemoney on the cost of the card?

(b) Cost without discount = £20 + £18 = £38b)

No card

1 trip

1 × £38 = £38

2 × £38 = £76

3 × £ 38 = £114

With card

£40 + (1 × £21.80) = £61.80

2 trips 2 × £38 = £76 £40 + (2 × £21.80) = £83.60

3 trips 3 x £38 = £114 £40 + (3 × £21.80) = £105.40

3 trips before the discount card starts being cheaper than regular price

3 0
3 years ago
Please help solve with work
sergey [27]
X-4y=28
8x+4y=8
---------------
9x=36→x=4
4-4y=28→y=6
—-----------------------------------------------------------
y-5=x
4(y-5)-y=4
4y-20-y=4
3y=24
y=8
x8-5=3
4 0
3 years ago
catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts. A banquet committee is to select 7 ​appetizers, 8 main​
guapka [62]

Answer:  The required number of ways is 46200.

Step-by-step explanation:  Given that a catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts.

A banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts.

We are to find the number of ways in which this can be done.

We know that

From n different things, we can choose r things at a time in ^nC_r ways.

So,

the number of ways in which 7 appetizers can be chosen from 8 appetizers is

n_1=^8C_7=\dfrac{8!}{7!(8-7)!}=\dfrac{8\times7!}{7!\times1}=8,

the number of ways in which 8 main courses can be chosen from 11 main courses is

n_2=^{11}C_8=\dfrac{11!}{8!(11-8)!}=\dfrac{11\times10\times9\times8!}{8!\times3\times2\times1}=165

and the number of ways in which 4 desserts can be chosen from 7 desserts is

n_3=^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!\times3\times2\times1}=35.

Therefore, the number of ways in which the banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts is given by

n=n_1\times n_2\times n_3=8\times165\times35=46200.

Thus, the required number of ways is 46200.

7 0
4 years ago
Select the correct answer.
Debora [2.8K]
As the risk of an investment decreases, so does the potential for its return
6 0
3 years ago
Read 2 more answers
8/11 divide by 3/5<br><br> find the quotient
AURORKA [14]

Answer:

=40/33

8/11 divide 3/5 -> 8/11 x 5/3 and multiply 8x5 then 11x3 then u got your answer! :)

Hope this worked :D

8 0
3 years ago
Read 2 more answers
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