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2 years ago

Help These are inscribed angles I think

Mathematics

1 answer:

MrRa [10]2 years ago

5 0

Answer:

13) 142° (intercepted arc is twice the inscribed angle)

14) 23° (inscribed angles is half of the intercepted arc)

15) Since RT is a diameter and arcRS = 46°, then arcST = 180° - 46 = 134°

16) Using the result of #13, arcRQ = 180° - 142° = 38°

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What is the dilation for the transformation below?

lys-0071 [83]

Answer:

The dialation is of 2

Step-by-step explanation:

QR = 3 units

QS = 2 units

RS = 2 units

Q'R' = 6 units

Q'S' = 4 units

R'S' = 4 units

So as you can see, this triangle has doubled it's size, since it has been multiplied by 2.

Hope this helped!

Have a supercalifragilisticexpialidocious day!

4 0

2 years ago

What the slope of (-15,9) (-10,3)

sweet [91]

The slope is - 3/10

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3 years ago

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I don't understand this problem on my homework can explain to me how to solve it?

user100 [1]

So he worked 30.5 hours, he just didn't write some of the hours down. To find the hours he didn't write down, you subtract 8+ $7\frac14$ + $8\frac12$ from 30.5, which equals $6 \frac14$

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3 years ago

Which results only in a horizontal compression of y=1/x by a factor of 6

yan [13]

we are given

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and we get

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so, option-A.........Answer

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2 years ago

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A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago