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dsp73
3 years ago
12

Simplify 11a +(4a-4)​

Mathematics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

Step-by-step explanation:

15a-14

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It should go like this:

(4z + 3) (3z - 4) / (3z - 4) (z + 2)

Then you just cancel out (3z -4) and (3z -4), and the final simplified form of this polynomial is:

(4z +3) / (z + 2)
4 0
4 years ago
Julia is the president of the technology Club at school she has $568 to buy as many DVD players as she can for $75 each how many
goblinko [34]
You need to divide the total amount by $75, to see how many DVD players she could buy with the $568.
568÷75= 7 remaining 43 
This means that Julia can buy 7 DVD players, and will have $43 extra dollars left over.
5 0
3 years ago
Write each Percent as a decimal and as a mixed number or fraction in simplest form
sveticcg [70]

p\%=\dfrac{p}{100}\\\\225\%=\dfrac{225}{100}=2\dfrac{25}{100}=2.25=2\dfrac{1}{4}\\\\550\%=\dfrac{550}{100}=\dfrac{55}{10}=5\dfrac{5}{10}=5.5=5\dfrac{1}{2}\\\\0.8\%=\dfrac{0.8}{100}=\dfrac{0.8\cdot10}{100\cdot10}=\dfrac{8}{1,000}=0.008=\dfrac{1}{125}\\\\0.06\%=\dfrac{0.06}{100}=\dfrac{0.06\cdot100}{100\cdot100}=\dfrac{6}{10,000}=0.0006=\dfrac{3}{5,000}

6 0
3 years ago
What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?
Marrrta [24]

C

the sum to infinity of a geometric series  = \frac{a}{1-r}

where a is the first term and r the common ratio

= \frac{81}{\frac{1}{3} } = 81 × 3 = 243



4 0
3 years ago
he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
3 years ago
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