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AysviL [449]
3 years ago
5

Please I need some answers

Mathematics
2 answers:
shutvik [7]3 years ago
5 0

Answer:

i) 13    ii) 25   iii) 20     iv) 34

Step-by-step explanation:

they are all exponents of 2 so you multiply each number by itself once.

i) 2x2 +  3x3

ii) 3x3 + 4x4

iii) 2x2 + 4x4

iv) 3x3 + 5x5

(next time you can use a calculator lol)

kolezko [41]3 years ago
3 0

Step-by-step explanation:

1)4+9 = 13

2)9+16 = 25

3)4+16 = 18

4)9+25 = 34

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A recent report claimed that 13% of students typically walk to school.8 DeAnna thinks that the proportion is higher than 0.13 at
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Can someone explain thoroughly how I am meant to solve this?
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Answer:

I got x<equal to -11 or x>equal to4

Step-by-step explanation:

x^2+15x+44=0

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x+4=0 or x+11=0(Set factors equal to 0)

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Check intervals in between critical points. (Test values in the intervals to see if they work.)

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−11<x<−4(Doesn't work in original inequality)

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Pls help on #7 I will give brainiest ! 7th grade level
valentina_108 [34]

Answer:

50

Step-by-step explanation:

To solve this problem you multiply 6 by 100 and then divide the total by 12 as follows: (6 x 100) / 12

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Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

Since is a one right tailed test the p value would be:  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

6 0
3 years ago
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