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3 years ago

Plz answer for 20 pts

Mathematics

2 answers:

Marina86 [1]3 years ago

6 0

Answer:

$6) \: \: \sqrt{3} + 4 \sqrt{3} \: = \: 5\sqrt{3}$

$7) \: \: 3\sqrt{5}+ 6\sqrt{45} = 3\sqrt{5}+6\sqrt{9} \! \cdot \! \sqrt{5}$

$= 3\sqrt{5} + 6\! \cdot \! 3 \! \cdot \! \sqrt{5}=3 \sqrt{5} + 18 \sqrt{5} = 21\sqrt{5}$

KATRIN_1 [288]3 years ago

5 0

Answer to Problem 1: 5√3

Answer to Problem 2: 21√5

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HELP ME WITH THIS (a+5)(b-3)

jonny [76]

Answer:

The answer is ab-3a+5b-15

Step-by-step explanation:

apply de disruptive property by multiplying each term a+5by each term b-3

Hope this helps :)

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3 years ago

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Chinedu sat down to do his homework, which included 30 math problems. He solved 2 problems each minute. Let f(n) be the number o

Katen [24]

Answer:

Step-by-step explanation:

If he is solving 2 problems per minute and 8 minutes have passed, he has solved

2*8 = 16 problems

There are 30 problems and he has solved 16

30-16 = 14

There are 14 problems left to solve at the start of the 9th minute

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4 years ago

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330% of 21 is what number?

kotykmax [81]

Answer: 69.3

Step-by-step explanation:

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4 years ago

From a bowl containing five red, three white, and seven blue chips, select four at random and without replacement. Compute the c

snow_lady [41]

Answer:

The probability is $\frac{5}{9}$

Step-by-step explanation:

Let A be the event of one red, zero white, and three blue chips,

And, B is the event of at least three blue chips,

Since, A ∩ B = A (because If A happens that it is obvious that B will happen )

Thus, the conditional probability of A if B is given,

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$

Now, red chips = 5,

White chips = 3,

Blue chips = 7,

Total chips = 5 + 3 + 7 = 15

Since, the probability of one red, zero white, and three blue chips, when four chips are chosen,

$P(A)=\frac{^5C_1\times ^3C_0\times ^7C_3}{^{15}C_4}$

$=\frac{5\times 35}{1365}$

$=\frac{175}{1365}$

$=\frac{5}{39}$

While, the probability that of at least three blue chips,

$P(B)=\frac{^8C_1\times ^7C_3+^8C_0\times ^7C_4}{^{15}C_4}$

$=\frac{8\times 35+35}{1365}$

$=\frac{315}{1365}$

$=\frac{3}{13}$

Hence, the required conditional probability would be,

$P(\frac{A}{B})=\frac{5/39}{3/13}$

$=\frac{65}{117}$

$=\frac{5}{9}$

4 0

3 years ago

The scores on a math test are normally distributed with a mean of 74 and a standard deviation of 8. The test scores range from 0

hram777 [196]

Answer:

The Estimate the number of students who took the scores between 82 and 98 = 16

Step-by-step explanation:

Explanation:-

Given data The scores on a math test are normally distributed with a mean μ = 74

standard deviation of Population

S.D (σ) = 8

Let 'x' be the random variable of Normal distribution

case(i):- when x = 82

$Z = \frac{x-mean}{S.D}$

$Z = \frac{82-74}{8} = 1$

case(ii):- when x = 98

$Z = \frac{x-mean}{S.D}$

$Z = \frac{98-74}{8} = 3$

The probability that test scores between 82 and 98.

P(82≤x≤98) = P(1≤z≤3)

= P(z≤3) - P(z≤1)

= 0.5+A(3)-(0.5+A(1))

= A(3) -A(1)

= 0.4986 - 0.3413

= 0.1573

Final answer:-

The Estimate the number of students who took the scores between 82 and 98

= 100 X 0.1573 = 15.73 ≅16

7 0

3 years ago