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3 years ago

Use the following function rule to find f(12). f(x)=10x-4, f(12)=

Mathematics

2 answers:

DaniilM [7]3 years ago

4 0

f(x)=10x-4, f(12)= 116

Novay_Z [31]3 years ago

3 0

Answer:

116

Step-by-step explanation:

Just replace x with 12 so:

f(12) = 10(12) - 4

Brainly plz :))

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A $1,287.99 TV was marked down at a 12% discount. What is the discounted price (or the reduced selling price)?

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2 years ago

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Least common denominator of 1/10 and 3/5

Alisiya [41]

Answer:

It's 10

Step-by-step explanation:

10 can be divided by five and 5 can also be divided by five and it's the least number that can be divided by five.

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3 years ago

Which calculation should be used to calculate s9 for the arithmetic sequence an=3n-1

Ganezh [65]

Answer: Choice A

S9 = (9/2)*(2+26)

===============================================

The formula is

Sn = (n/2)*(a1+an)

where

Sn = sum of the first n terms (nth partial sum)

n = number of terms

a1 = first term

an = nth term

In this case,

n = 9

a1 = 2 (plug in n = 1 into the formula an = 3n-1 and simplify)

an = a9 = 26 (plug n = 9 into the formula an = 3n-1 and simplify)

So,

Sn = (n/2)*(a1+an)

S9 = (9/2)*(2+26)

will help us find the sum of the first 9 terms of this arithmetic sequence

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3 years ago

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A) Convert 23ten to binary.

WITCHER [35]

Answer:

answer : 10111

Step-by-step explanation:

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3 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago