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3 years ago

Please solve. 6(3 + 8) - 4/5

Mathematics

2 answers:

Rudik [331]3 years ago

7 0

Answer:

$\Large \boxed{\frac{326}{5}}$

Step-by-step explanation:

$\displaystyle 6(3 + 8) - \frac{4}{5}$

Solving brackets.

$\displaystyle 6(11) - \frac{4}{5}$

$\displaystyle 66 - \frac{4}{5}$

Subtract.

$\displaystyle \frac{66}{1} - \frac{4}{5}$

$\displaystyle \frac{66 \times 5}{1 \times 5} - \frac{4}{5}$

$\displaystyle \frac{330}{5} - \frac{4}{5}$

$\displaystyle \frac{330-4}{5}$

$\displaystyle \frac{326}{5}$

Gre4nikov [31]3 years ago

3 0

G'Day mate! Here's the answer:

53 1/3

1. 6(3 + 6) - 4/5

2. 6 x 3 is 18, and 6 x 8 is 48

3. 18 plus 48 is 66

4. 66 - 4/5 = 266/5

5. 266/5 as mixed number is 53 1/5

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Two weeks in a row, the golf course hosts a group of golfers. The second week had 10 more golfers than the first week. Use the d

Trava [24]

Answer:

Option A.

Step-by-step explanation:

From the first dot plot the given data set for week 1 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69), (70, 70, 72, 72, 72), (75, 75, 76, 78, 78), (80, 80, 80, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{72+75}{2}=73.5$

$Mean=\frac{\sum x}{n}=\frac{1491}{20}=74.55$

From the second dot plot the given data set for week 2 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 85, 86, 86, 86, 88, 88, 88, 88, 89, 89, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69, 70, 70), 72, (72, 72, 75, 75, 76, 78, 78), (80, 80, 80, 85, 86, 86, 86), 88, (88, 88, 88, 89, 89, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{78+80}{2}=79$

$Mean=\frac{\sum x}{n}=\frac{2364}{30}=78.8$

The range for Week 1 is equal to the range for Week 2.

The median for Week 2 is more than the median for Week 1.

The mean for Week 2 is more than the mean for Week 1.

Therefore, the correct option is A.

6 0

3 years ago

An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago

A ramp leading into a building makes a 15 degree angle with the ground. The end of the ramp is 10 feet from the base of the buil

stiks02 [169]

Answer:

The length of the ramp is x = AC = 10.35 feet

Step-by-step explanation:

From the ΔABC

AC = x = length of ramp

BC = 10 feet

$\theta = 15$ °

$\cos 15 = \frac{BC}{AC}$

$\cos 15 = \frac{10}{x}$

x = 10.35 feet

Therefore the length of the ramp is x = AC = 10.35 feet

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4 years ago

Is the graph nonlinear or linear??

SVETLANKA909090 [29]

Answer: I believe it's nonlinear

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3 years ago

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An advertisement says that 4 out of 5 doctors prefer a certain product. what percent of doctors do not like that product

Nina [5.8K]

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$\dfrac{1}{5} \times 100\% = 20\%$

Answer : 20% of doctors do not like that product.

Hope this helps. - M

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3 years ago

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