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2 years ago

Please help i keep getting this wrong

Mathematics

2 answers:

svp [43]2 years ago

6 0

Answer:

(5x+10y)/(x+2y)

Step-by-step explanation:

forsale [732]2 years ago

6 0

You find greatest common factor for both the numerator part and the denominator part.

First look at numerator
5x^2 10xy

What is the greatest common factor?

5 and x

So
5x (x + 2y)

[you can multiply them back and see that you get the initial equation. Also, cant simplify x+2y any further so this is correct]

Now the denominator:
X^2 +2xy

Again, find GCF.
It is only x

So,

X(x+2y)

Now what do we have?

5x(x+2y)/ x(x+2y)

You can see that x+2y is present both in numerator snd in denaminator, so you can cancel them out.
You are left with.

5x/x

You can cancel x’s too and you are left with

5

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3 tenths + 4 Tenths = __0.7____ tenths

vovangra [49]

Answer:

7 tenths

Step-by-step explanation:

3 tenths + 4 tenths = 3/10 + 4/10 = 0.3 + 0.4 = 0.7

= 7 tenths

6 0

2 years ago

What is the Composite Surface Area? Need help asap is due today.

jok3333 [9.3K]

I’m really not sure but 216 because bass multiplied by height multiplied by width (6*6*6)

4 0

2 years ago

At an annual salary of $55,000 per year, what is the amount of semimonthly paychecks?

Amiraneli [1.4K]

Semi monthly means 2 times per month
assume that pay is at beginning and middle of month (doesn't matter, but makes it easier for me)

so 12 months
2 times per month
12 times 2=24
divide 55000 by number of months
55,000/24=2291.67

so the amount per paycheck, assuming equally distributed amounts throughout the paychecks, is $2291.67

4 0

3 years ago

if in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99%

Dahasolnce [82]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.

If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?

Given Information:

standard deviation = σ = 30 hours

confidence level = 99%

Margin of error = 6 hours

Required Information:

sample size = n = ?

Answer:

sample size = n ≈ 165

Step-by-step explanation:

We know that margin of error is given by

Margin of error = z*(σ/√n)

Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size

√n = z*σ/Margin of error

squaring both sides

n = (z*σ/Margin of error)²

For 99% confidence level the z-score is 2.576

n = (2.576*30/6)²

n = 164.73

since number of bulbs cannot be in fraction so rounding off yields

n ≈ 165

Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.

3 0

2 years ago

Please help me answer thanks.

Dmitry_Shevchenko [17]

Answer:

see explanation

Step-by-step explanation:

Given that M is directly proportional to r³ then the equation relating them is

M = kr³ ← k is the constant of proportion

To find k use the condition when r = 4, M = 160, thus

160 = k × 4³ = 64k ( divide both sides by 64 )

2.5 = k

M = 2.5r³ ← equation of variation

(a)

When r = 2, then

M = 2.5 × 2³ = 2.5 × 8 = 20

(b)

When M = 540, then

540 = 2.5r³ ( divide both sides by 2.5 )_

216 = r³ ( take the cube root of both sides )

r = $\sqrt[3]{216}$ = 6

5 0

3 years ago