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3 years ago

7

johnny saves all of his nickles and dimes in a jar. One day he counted them and found that there were 73 coins worth $4.55. How

many nickels and dimes were in the jar?

1 answer:

juin [17]3 years ago

7 0

Ummm I'm not sure but let me do the math and I'll tell u :)

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In a study, 36% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop

steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.

i.e. $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

Formula : $P(x=r)=^nC_r p^r q ^ {n-r}$

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

$P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=0.390748$

Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907

5 0

2 years ago

1. You are designing a rectangular playground. On your scale drawing, the vertices of the rectangle are (6, 3), (6, 5), and (8,

kramer

(8,5) because (6,5) is distance 2 from (6,3). So (8,5) will distance 2 from (8,3)

5 0

2 years ago

Read 2 more answers

The coordinates of F are (10,5). The coordinates of G are (-8,11). Find the midpoint of F G. Write your answer as an ordered pai

vova2212 [387]

Answer: (1, 8)

Step-by-step explanation:

6 0

2 years ago

Subtract rational expression. <br> Show work please.

yan [13]

Given:

$$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}$

To find:

The simplified rational expression by subtraction.

Solution:

Let us factor $x^2-1$ . It can be written as $x^2-1^2$ .

$x^2-1^2=(x-1)(x+1)$ using algebraic identity.

$$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x+4}{x-1}-\frac{5}{(x+1)(x-1)}$

LCM of $x-1,(x+1)(x-1)=(x+1)(x-1)$

Make the denominators same using LCM.

Multiply and divide the first term by (x + 1) to make the denominator same.

$$=\frac{(x+4)(x+1)}{(x-1)(x+1)}-\frac{5}{(x-1)(x+1)}$

Now, denominators are same, you can subtract the fractions.

$$=\frac{(x+4)(x+1)-5}{(x-1)(x+1)}$

Expand $(x+4)(x+1)-5$ .

$$=\frac{x^2+4x+x+4-5}{(x-1)(x+1)}$

$$=\frac{x^{2}+5 x-1}{(x-1)(x+1)}$

$$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x^{2}+5 x-1}{(x-1)(x+1)}$

6 0

2 years ago

3) A box of chocolates contains four milk

ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

huh

3 0

3 years ago

Read 2 more answers