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2 years ago

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

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

So, 96% confidence interval for the population mean, $\mu$ is ;

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

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

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

= [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