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balu736 [363]
3 years ago
8

Evaluate the expression when c= -5. c^2+ 6c-9

Mathematics
1 answer:
sesenic [268]3 years ago
3 0

Answer:

<h2><em><u>-64</u></em></h2>

Explanation:

c² + 6c - 9

  • Insert -5 For 'c'

-5² + 6(-5) - 9

  • Follow P.E.M.D.A.S.
  • 5² = 25

= -25 + 6(-5) - 9

  • Multiply: 6(-5) = -30

= -25 - 30 - 9

  • Simplify
<h2><em><u>-64</u></em></h2>
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4 0
3 years ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is
Leto [7]

Answer:

t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827    

The degrees of freedom are given by:

df=n-1=23-1=22  

And the p value taking in count that we have a bilateral test we got:

p_v =2*P(t_{(22)}>0.827)=0.417  

Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

Step-by-step explanation:

Information given

\bar X=435 represent the mean for the weight

s=\sqrt{841}=29 represent the sample standard deviation

n=23 sample size  

\mu_o =430 represent the value that we want to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

System of hypothesis

We are trying to proof if the filling machine works correctly at the 430 gram setting, so then the system of hypothesis for this case are:

Null hypothesis:\mu = 430  

Alternative hypothesis:\mu \neq 430  

In order to cehck the hypothesis the statistic for a one sample mean test is given by

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we have this:

t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827    

The degrees of freedom are given by:

df=n-1=23-1=22  

And the p value taking in count that we have a bilateral test we got:

p_v =2*P(t_{(22)}>0.827)=0.417  

Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

8 0
3 years ago
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