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True [87]
3 years ago
14

2+3x34 what is the answer in unit form

Mathematics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

3 x 34 = 102 + 2 = 104 is your answer just do the equation

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Triangle abc is translated 5 units to the right and 4 units up to create triangle A’B’C .
eimsori [14]

Answer:

This cannot be solved unless I have the coordinate for triangle "abc"

Step-by-step explanation:

Once I have the coordinates for triangle "abc", I can solve to find out where triangle "A'B'C" is

8 0
2 years ago
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
How can i solve this? please help.
iogann1982 [59]

Answer:SEARCH NYO SA GOGGLE

Step-by-step explanation:PLZZ

6 0
2 years ago
Find the complete factored form of the polynomial 9ab^5+6a^3b^4
notsponge [240]
9ab^5+6a^3b^4 

GCF=3ab^4 

3ab^4( \dfrac{9ab^5}{3ab^4} +  \dfrac{6a^3b^}{3ab^4} ) 

3ab^4(3b+2a^2)
6 0
3 years ago
The least amount of wrapping paper needed to wrap a cube-shaped gift is 150 square inches. How long is one side of the gift?
devlian [24]


Isn't the only thing you have to do is since a cube has 6 sides than you would have to do 150 divided by 6. If I am correct.


5 0
3 years ago
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