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Zarrin [17]
3 years ago
8

The inverse of F(x) is a function.

Mathematics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

no it is not a function becouse you didn't add the other value

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The subtraction property of equality states that the two sides of an
yawa3891 [41]

Answer:

False

Step-by-step explanation:

The answer is in its name. Its a property of equality meaning it makes sure that both sides are equal to each other

3 0
2 years ago
A few times a week, Scott treats himself to a small latte from his favorite coffee shop. Each latte costs $3.55. If Scott bought
iogann1982 [59]

Answer:

Scott spent $53.25 on lattes last month.

Step-by-step explanation:

To find how much Scott spent in total, multiply the number of lattes bought by the cost of each latte:

15 × 3.55

53.25

5 0
3 years ago
An office supply store recently sold a black printer ink cartridge for $14.99 and a color printer ink cartridge for $20.99. At s
nevsk [136]

Answer:

B = 31, C = 11

Step-by-step explanation:

Let B be the number of Black printer ink cartridges and C be the number of color printer ink cartridges.

42 were sold in total, hence,

B+C= 42 ------------(1)

Also, its given that

14.99 B+20.99C= 695.58 ------(2)

Multiplying (1) by 20.99,

14.99 B+14.99 C= 629.58 ------(3)

Now, subtracting (3) from (2),

14.99 B+20.99C= 695.58

14.99 B+14.99 C= 629.58

------------------------------------

6 C = 66

C= 66/6

C = 11

We know, B+ C= 49

Hence B = 42- 11 = 31

B = 31

Hence, number of Black printer ink cartridges = 31 and color printer ink cartridges = 11 .

4 0
3 years ago
Takuya's parents gave him \$100$100dollar sign, 100 as a birthday gift. Since he loves board games, he spent \$20$20dollar sign,
Delicious77 [7]

Answer:

The amount of money that remained from Takuya's present during month n is expressed as Tn = 120-20n and the sequence is an ARITHMETIC SEQUENCE.

Step-by-step explanation:

If Takuya's parents gave him $100 as a birthday gift and she spent $20 each month on board games until his money ran out, this means he keeps spending $20 every month and his money keep reducing by the same amount each month until his money ran out, the following can be inferred;

initial amount = $100

If he spends $20 each month,

balance at the end of 1st month = $100-$20 = $80

balance at the end of 2nd month = $80-$20 = $60

balance at the end of  3rd month = $60-$20 = $40 and so on

The sequence formed by his balances is $100, $80, $60, $40...

Since the amount keep reducing by the same value i.e $20, then the sequence formed is an ARITHMETIC SEQUENCE.

The nth term of an arithmetic sequence is expressed as Tn = a+(n-1)d

a is the first term of the sequence = 100

d is the common difference = 80-100 = 60-80 = 40-60 = 20

n is the number of terms

Substituting the given values in the formula we have;

Tn = 100+(n-1)*-20

Tn = 100+(-20n+20)\\Tn = 100-20n+20\\Tn = 120-20n

The amount of money that remained from Takuya's present during month n is expressed as Tn = 120-20n and the sequence formed is an ARITHMETIC SEQUENCE

6 0
4 years ago
The lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length
snow_lady [41]

Answer: About 99.74% of births would be expected to occur within 24 days of the mean pregnancy length.

Step-by-step explanation:

Complete question is attached below.

Given: The lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length with a standard deviation of 8 days.

i.e. \sigma= 8

let X denotes the random variable that represents the lengths of pregnancy.

The probability of births would be expected to occur within 24 days of the mean pregnancy​ length:

P(\mu-24

= 2(0.9987)-1\ \ \ [\text{ By z-table}]\\\\=0.9974

=99.74%

Hence, about 99.74% of births would be expected to occur within 24 days of the mean pregnancy length.

4 0
3 years ago
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