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3 years ago

Find the output, k, when the input, t, is –7. k = 10t -19

Mathematics

1 answer:

kupik [55]3 years ago

7 0

Answer:

k = - 89

Step-by-step explanation:

To calculate the output, substitute t = - 7 into the equation for output, that is

k = 10(- 7) - 19 = - 70 - 19 = - 89

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What is the solution of √-4x=100?<br> A) x = –2500<br> B) x = –50<br> C) x = –2.5<br> D) no solution

yawa3891 [41]

First we raise both members of the equality to the square:
(√-4x) ^ 2 = (100) ^ 2
Then, we solve:
-4x = 10000
We cleared x in the last instance
x = 10000 / (- 4)
x = -2500
answer:
the solution of √-4x=100 is
A) x = –2500

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3 years ago

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A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed f

Olin [163]

Answer:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the frequencies

H1: There is a difference in the frequencies

The level of significance assumed for this case is $\alpha=0.01$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are:

Value 1 2 3 4 5 6

Frequency 26 32 44 37 27 34

And the expected values are for this case the same $E_i = \frac{200}{6}= 33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

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4 years ago

Clark made four of his truck payments late and was fined four late fees. The total change to his savings from late fees was-$40.

luda_lava [24]

You would do 40 ÷ 4 which would be 10. so it was $10 per late fee

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Hitman42 [59]

Answer: (740)((1+0.0065)^132-1)/(0.0065)(1+0.0065)^132

Step-by-step explanation:

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3 years ago