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3 years ago

Can some one help, step bye step please

Mathematics

2 answers:

il63 [147K]3 years ago

4 0

Answer:

$14.23

Step-by-step explanation:

Multiply the original value by 1.035 to get your answer.

Thepotemich [5.8K]3 years ago

3 0

Let x be the initial salary

x=13.75

Let y be the new salary

y=x+x(3.5%)

y=13.75+13.75(0.035)

y=13.75+0.48=14.23

New hourly salary=14.23$

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Celine ate brunch at a restaurant. The bill came to $72.80. If she left a 15% tip, what was the total cost of her brunch?

DedPeter [7]

Answer:

$83.69

Step-by-step explanation:

Given the following

Initial Cost of the brunch = $72.80

Percentage tip = 15%

Tip paid = 15% of $72.60

Tip paid = 0.15 * 72.60

Tip paid = $10.89

Total cost of brunch = $72.80 + $10.89

Total cost of brunch = $83.69

Hence the total cost of her brunch is $83.69

7 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

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

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$