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Vanyuwa [196]
2 years ago
15

When f(x)=5x^2-2x+5, evaluate f(-3)

Mathematics
1 answer:
patriot [66]2 years ago
8 0

Answer:

f(-3)=56

Step-by-step explanation:

When evaluating a specific input for a function, you just plug that input in for x.

So in this case, f(-3)=5(-3)^2-2(-3)+5

f(-3)=5(-3)^2-2(-3)+5

f(-3)=5(9)+6+5

f(-3)=45+6+5

f(-3)=56

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Kenneth has 2,000$ to invest
Katarina [22]

The calculations for 2, 4, 8 year can be derived as follows,

For simple interest:

                   S.I = \frac{P*T*R}{100}

P= principle

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For 2 year;

            S.I = \frac{2000*2*5.6}{100}=  $224

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For 4 year;

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3 years ago
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2 years ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 402 gram setting. It is
lidiya [134]

Answer:

The value of t test statistics is -2.25.

Step-by-step explanation:

We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 402 gram setting. It is believed that the machine is under filling the bags.

A 25 bag sample had a mean of 393 grams with a standard deviation of 20.

<u><em>Let </em></u>\mu<u><em> = mean filling of bags by the machine.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 402 gram     {means that the the machine is not under filling the bags}

Alternate Hypothesis, H_A : \mu < 402 gram     {means that the the machine is under filling the bags}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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

where, \bar X = sample mean bag filling = 393 grams

            s = sample standard deviation = 20 grams

            n = sample of bags = 25

So, <u><em>test statistics</em></u>  =  \frac{393-402}{\frac{20}{\sqrt{25} } }  ~ t_2_4

                               =  -2.25

The value of t test statistics is -2.25.

5 0
3 years ago
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