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Andre45 [30]
2 years ago
11

Which statement is not used to prove that triangle ABC is similar to triangle ADC?

Mathematics
1 answer:
yarga [219]2 years ago
6 0

Answer:

what are the statements we pick from?

Step-by-step explanation:

most likely look out for angles and lengths that go against the established ratio

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When a number is added to six times its square, the result is twelve. Find the number.
7nadin3 [17]
Let's represent our number as x.
x is being added to 6 * x², and it equals 12
x + 6x^{2}=12\\6x^{2}+x-12=0
Factor:
6x^{2}+x-12=0\\(2x+3)(3x-4)=0
Solve for x:
(2x+3)=0\\2x=-3\\x=\frac{-3}{2}\\\\(3x-4)=0\\3x=4\\x=\frac{4}{3}

Therefore, x = -3/2 and 4/3.
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Natalka [10]

Answer:

3 is b

4 is also b

Step-by-step explanation:

6 0
3 years ago
5(x+3)-5<br> simplify this.
nignag [31]

Answer:

5x+10

Step-by-step explanation:

Distribute:

=(5)(x)+(5)(3)+−5

=5x+15+−5

Combine Like Terms:

=5x+15+−5

=(5x)+(15+−5)

=5x+10

8 0
3 years ago
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

5 0
3 years ago
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