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Nesterboy [21]
2 years ago
14

Francisco is saving money in an account. At the beginning of the year, he has $56 in savings

Mathematics
2 answers:
aliina [53]2 years ago
5 0

208 would be the  anser i think

earnstyle [38]2 years ago
3 0
It’s 208$ dollars with out adding the 56$ dollars if you add the 56$ dollars its 264$ dollars

Explanation do 52 times 4 equals 208 plus 56 and you get 264
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Compute​ P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate thi
stich3 [128]

Answer:

The answers to the questions are;

A) P(X) where x = 12 is equal to 3.55 ×10⁻²

B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²

C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10

E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.

Step-by-step explanation:

To solve the question, we note that

n = 58

p = 0.3

x = 12

Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability

A) We are to use the binomial probability formula to find P(X)

Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶

= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²

B) Using the standard distribution table we have

the z score for x = 12 given as z = \frac{x-\mu}{\sigma} = -1.55

From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by

\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }

where:

σ = Standard deviation = \sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is \frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}

C)  The probabilities differ by 3.55 ×10⁻² -  3.453 ×10⁻² =  9.7×10⁻⁴

D) The normal distribution be used to approximate this probability because     n·p and n·p·(n-p) ≥ 5

E)   f_i represents the area under the curve towards left of the ith data observed in a normally distributed population.

Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value  

5 0
2 years ago
Helppp pleaseeee!!!!
Savatey [412]

Answer:

Initial value = 4

Growth factor = 1.6

Step-by-step explanation:

Function representing the exponential growth is given by,

f(x) = Initial value(1 + growth rate)ˣ

Here x = time or duration for growth

(1 + growth rate) = Growth factor

Given function is,

f(x) = 4(1.6)ˣ

By comparing both the functions,

Initial value = 4

Growth factor = 1.6

4 0
3 years ago
What is the range of the function f(x) = 4x + 9, given the domain D = {-4, -2, 0, 2}? please help
neonofarm [45]
Since this is a linear function, filling in the minimum and maximum of the domain is sufficient.

f(-4) = -16 + 9 = -7
f(2)= 8 + 9 = 17

So the range of the function (given the domain) :
R = {-7, 17}
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Which letters have 1-fold reflectional symmetry?
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capital A

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lower case and upper case o

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3 years ago
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