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2 years ago

12

Ms. Willett joins Planet Fitness which charges a monthly membership of $19.99. How much will Ms. Willett spend in all if she con

tinues her
membership for 1.5 years?

2 answers:

zimovet [89]2 years ago

7 0

First, you have to know that 1.5 years equals 18 months. Next, just multiply $19.99 by 18. This will give you a final answer of $359.82. Hope this helps!

Rashid [163]2 years ago

6 0

Answer:

$29.985

Step-by-step explanation:

19.99 x 1.5 = 29.985

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Calculate 65 L to quarts. Final answer round two places

lakkis [162]

Step-by-step explanation:

The approximate equation to convert from liters to quarts is : x • 1.056688.

We enter 65 for x and multiply.

65 • 1.056688 = 68.68472

We round that to the nearest 100th to get our final answer.

Our final answer: 68.68 liters

7 0

3 years ago

1) Find the values of 'a' and 'b' for the given parallelogram *

nataly862011 [7]

Answer:

first option

Step-by-step explanation:

In a parallelogram consecutive angle are supplementary, sum to 180, so

2a + 122 = 180 ( subtract 122 from both sides )

2a = 58 ( divide both sides by 2 )

a = 29

---------------

In a parallelogram the opposite sides are congruent, then

b + 6 = 10 ( subtract 6 from both sides )

b = 4

5 0

2 years ago

Find the range, mean, and standard deviation of the data set (3, 6, 9, 11, 15, 16) The range is 13 (Type a whole number.) The me

iris [78.8K]

Answer: The range is 13. The mean is 10. The standard deviation is 4.62 .

Step-by-step explanation:

The given data : 3, 6, 9, 11, 15, 16

Total number of data values : n = 6

The mean of data is given by :-

$\overline{x}=\dfrac{\sum^6_{i=1}x_i}{n}\\\\\Rightarrow\overline{x}=\dfrac{60}{6}=10$

The standard deviation is given by :-

$\sqrt{\dfrac{1}{n}(\sum^6_{i=1}(x_i-\overline{x})^2)}\\\\=\sqrt{\dfrac{1}{6}(\sum^6_{i=1}(x_i-10)^2)}\\\\=\sqrt{\dfrac{1}{6}\times(49+16+1+1+25+36)}=4.61880215352\approx4.62$

The range of the data : Maximum value -Minimum value

$=16-3=13$

6 0

2 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

2 years ago

Sam invests $6000 in two different accounts. The first account paid 12 %, the second account paid 7 % in interest. At the end of

Lunna [17]

Answer:

The amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

Step-by-step explanation:

Let x be the amount in the account at 12% interest

So, 6000-x is the amount in the second account at 7% interest

$SI = \frac{P \times T \times R}{100}$

First account: $SI=\frac{x \times 1 \times 12}{100}$

Second account : $SI =\frac{(6000-x) \times 1 \times 7}{100}$

We are given that At the end of the first year he had earned $590 in interest.

So, $\frac{x \times 1 \times 12}{100}+\frac{(6000-x) \times 1 \times 7}{100}=590\\x=3400$

So,the amount in the account at 12% interest is $3400

The amount in the second account at 7% interest =6000-x=6000-3400=2600

Hence the amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

8 0

3 years ago