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mixas84 [53]
2 years ago
11

HELP FAST DUE IN TEN MINUTES!!

Mathematics
1 answer:
goldenfox [79]2 years ago
5 0

Answer:

Someone is spending 80 cents daily from their bank account, starting out with $5.6

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Decrease 950 by 36%.
horsena [70]

Answer:

342

Step-by-step explanation:

950×0.36=342.....

6 0
3 years ago
Read 2 more answers
A store has clearance items that have been marked down by 60%. They are having a sale, advertising an additional 15% off clearan
astra-53 [7]

Answer: 34%

Step-by-step explanation:

Given

Price is marked down by 60% with an additional discount of 15% on the marked down price

Suppose x is the original price of the item

After marking down, it is

\Rightarrow x-x\times 0.6=0.4x

after additional discount it is

\Rightarrow 0.4x(1-0.15)=0.4x\times 0.85=0.34x

Effective discount on the item is

\Rightarrow \dfrac{x-0.34x}{x}\times 100=66\%

Thus, a person is paying 100-66=34\% of the original price

6 0
2 years ago
Consider the function f(x) = 10x^2 + 6x - 2 when x=3​
Savatey [412]

Answer:

106

Step-by-step explanation:

Substitute x = 3 into f(x) and evaluate , that is

f(3)

= 10(3)² + 6(3) - 2 = 10(9) + 18 - 2 = 90 + 18 - 2 = 106

5 0
3 years ago
A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen
tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the  average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

                         P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average desired retirement age = 55 years

            \sigma = sample standard deviation = 3.4 years

            n = sample of seniors = 101

            \mu = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.114 < t_1_0_0 < 2.114) = 0.96  {As the critical value of t at 100 degree

                                               of freedom are -2.114 & 2.114 with P = 2%}  

P(-2.114 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.114) = 0.96

P( -2.114 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X-2.114 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

<u>96% confidence interval for</u> \mu = [ \bar X-2.114 \times {\frac{s}{\sqrt{n} } } , \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ]

                                           = [ 55-2.114 \times {\frac{3.4}{\sqrt{101} } } , 55+2.114 \times {\frac{3.4}{\sqrt{101} } } ]

                                           = [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0
3 years ago
Consider this expression.
Oliga [24]
A
D
F
That is the correct answer
4 0
2 years ago
Read 2 more answers
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