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IRISSAK [1]
2 years ago
6

Can someone help me please?! :’)

Mathematics
1 answer:
zloy xaker [14]2 years ago
6 0

Answer: ##########################

Step-by-step explanation: YW

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TIMED TEST PLEASE HELP
xxMikexx [17]
1 ) first offer
total payments
375.76×12×4
=18,036.48
Interest paid
18,036.48−16,000
=2,036.48
Second offer
Total payments
390.61×12×4
=18,749.28
Interest paid
18,749.28−16,000
=2,749.28
Larry will save of taking 6% loan
2,749.28−2,036.48=712.8. .answer

2) credit card 1
Total payments
277.09×12
=3,325.08
Interest paid
3,325.08−3,000
=325.08
Credit card 2
Total payments
152.69×12×2
=3,664.56
Interest paid
3,664.56−3,000
=664.56
Susan will save
664.56−325.08
=339.48...answer

Hope it helps!
6 0
2 years ago
Read 2 more answers
Tickets to the zoo cost $12 for adults and $8 for children. The school has a budget of $240 for the field trip. An equation repr
Oxana [17]

Answer:

18 children

Step-by-step explanation:

Given

240 = 12x +8y

<em>Required (missing part of the question):</em>

<em>Number of children that will be able to go to the zoo if 8 adult tickets are purchased</em>

 

<em>The graph is missing. However, the question can be solved without the graph.</em>

<em />

From the question and equation, we can deduce the following:

  • x represents adults
  • y represents children

So, when 8 adult tickets were sold.

This means

x = 8

Substitute 8 for x in 240 = 12x +8y

240= 12 * 8 + 8y

240= 96+ 8y

Subtract 96 from both sides

240 - 96 = 96 - 96 + 8y

144 =  8y

8y = 144

Solve for y

y= \frac{144}{8}

y= 18

<em>This means that 18 children will be able to go</em>

5 0
2 years ago
What is the quotient StartFraction 15 p Superscript negative 4 Baseline q Superscript negative 6 Baseline Over negative 20 p Sup
zhannawk [14.2K]

Answer:

- \frac{3}{4} \times  \frac{p^{8} }{q^{3} }

Step-by-step explanation:

We have to find the quotient of the following division, \frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}.

Now, \frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}

= - \frac{3}{4} p^{[- 4 - (- 12)]} q^{[-6 - (- 3)]} {Since all the terms in the expression are in product form, so we can treat them separately}

{Since we know the property of exponent as \frac{a^{b} }{a^{c} } = a^{(b - c)}}

= - \frac{3}{4} p^{8} q^{-3}

= - \frac{3}{4} \times  \frac{p^{8} }{q^{3} } (Answer)

{Since we know, a^{-b} = \frac{1}{a^{b} }}

3 0
2 years ago
Read 2 more answers
you are to give a client one tablet labeled 0.15 mg and one labeled 0.025 mg. what is the total dosage of these two tablets
Ivenika [448]
Simply add the two dosages together, (0.15+0.025) and the answer is 0.175 :)
7 0
3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
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