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Leviafan [203]
3 years ago
5

The temperature in Waterloo was − 5°C at sunrise. By noon the temperature had risen by 12 Celsius degrees. What was the temperat

ure in Waterloo at noon?
Mathematics
2 answers:
allochka39001 [22]3 years ago
6 0
The answer is 7° Celsius.
posledela3 years ago
4 0
The answer would be 7 degrees celcius
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27^x = 9^x − 4 Question 1 options: x = 8 x = 4 x = −4 x = −8
Mariulka [41]

Answer:

x = -8

Step-by-step explanation:

Given the equation:

27^x=9^{x-4}

Note that

27=3^3\\ \\9=3^2,

then

27^x=(3^3)^x=3^{3x}\\ \\9^{x-4}=(3^2)^{x-4}=3^{2(x-4)}

Rewrite the equation as

3^{3x}=3^{2(x-4)},

then

3x=2(x-4)\\ \\3x=2x-8\\ \\3x-2x=-8\\ \\x=-8

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In this figure, lines a and b intersect a transversal, c.
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Answer: Alternate exterior angles

Step-by-step explanation:

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2 years ago
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Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 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.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

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

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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What number is 24% of 80
ch4aika [34]
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3 years ago
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Lucas deposits $4,000.00 for 6 years at 2% interest, compounded monthly. How much interest does he earn at the end of the 6 year
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Answer:

$12644.56

Step-by-step explanation:

one year = 12 month

4,000 * (1.02)^(6*12) = 16644.5615

16644.5615-4000 = 12644.5615

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2 years ago
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