Multiply and simplify: –5gh4 • (

In an election, 471 people voted. Candidate B received 2/3 of the votes. How many votes did Candidate B receive?

dezoksy [38]

Answer:

Step-by-step explanation:

471/3 * 2 = 314

5 0

4 years ago

Read 2 more answers

I need a answer to 77.5% of 13.270

den301095 [7]

Answer:

10.28425

Step-by-step explanation:

4 0

3 years ago

Rename 4/7 as a percent. Round to the nearest tenth of a percent if necessary.

enyata [817]

Answer:

Part 1) Option C. 57.1%

Part 2) Option B. 70.9%

Part 3) Option C. 66 2/3 %

Part 4) Option A. About 16

Part 5) Option A. About $12

Part 6) Option B. About $9

Part 7) Option B. x/50=70/100

Part 8) Option B. 25%

Part 9) Option D. 800

Part 10) $1,770

Step-by-step explanation:

Part 1) Rename 4/7 as a percent

we know that

To rename a number as a percent, multiply the number by 100

so

(4/7)*100=400/7=57.1%

Part 2) Rename .709 as a percent

we know that

To rename a number as a percent, multiply the number by 100

so

0.709*100=70.9%

Part 3) If 1/3 is equivalent to 33 1/3% what percent is equivalent to 2/3?

we know that

33 1/3%=(33*3+1)/3 %=100/3 %

By proportion

(100/3 %)/(1/3)=x/(2/3)

x=200/3 %=66 2/3 %

Part 4) Alice plays basketball. In one game she shoots 21 free throws and makes 81.25% of them. Estimate how many she makes

To find the exact value multiply the total shoots by the percentage

21*(81.25/100)=17.06

so

To find the estimate value

21 shoots is about 20

81.25% is about 80%

so

20*(80/100)=16

the answer is about 16

Part 5) Estimate 48% of 23.95

To find the exact value multiply the number by the percentage

23.95*(48/100)=11.50

To find the estimate value

23.95 is about 24

48% is about 50%

so

24*(50/100)=12

Part 6) The bill at a restraint is $58.50 and Mrs. Johnson wants to leave a 15% tip. Approximately how much money

To find the exact value multiply the bill by the percentage

so

$58.50*(15/100)=$8.78

To find the approximate value

$58.50 is about $60

$60*(15/100)=$9

Part 7) Which proportion can be used to solve the following percent problem? What is 50% if 100% is 70?

using proportion

x/50=70/100

x=70*50/100

x=35

Part 8) What percent of 50 is 12.5?

using proportion

50/100%=12.5/x%

x=100*12.5/50

x=25%

Part 9) 50 Is 6.25% of what number?

using proportion

50/6.25%=x/100%

x=100*50/6.25

x=800

Part 10) Chris works at Higgins’ Automart and is selling a new Shelby Mustang for $59,000. If he earns a commission of 3%, how much money does he earn?

we know that

$59,000 ------> represent the 100%

using proportion

Find how much money represent 3% of the commission

59,000/100%=x/3%

so

X=$59,000*(3/100)=$1,770

4 0

3 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

4 years ago

What is the answer 2( 4.8j + 4.57 ) ≤ - 19.06 – 18.6j

vesna_86 [32]

Answer:

j $\leq$ -1

Step-by-step explanation:

7 0

3 years ago

Multiply and simplify: –5gh4 • (–2g3h5)