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oee [108]
3 years ago
7

(12y2+17y-4)+9y2-13y+3)

Mathematics
1 answer:
Flura [38]3 years ago
7 0
(12y^2+17y-4)+(9y^2-13y+3)=12y^2+9y^2+17y-13y-4+3=\\\\=21y^2+4y-1
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what is the similarity ratio of the smaller to the larger similar cylinders? enter your answer as a:b​
Andre45 [30]

Answer:

2 : 3

Step-by-step explanation:

Given:

The two cylinders are similar. So, their corresponding dimensions must be in proportion.

\frac{\textrm{Radius of small cylinder}}{\textrm{Radius of larger cyclinder}}=\frac{12}{18}=\frac{2}{3}=2:3

\frac{\textrm{Height of small cylinder}}{\textrm{Height of larger cylinder}}=\frac{20}{30}=\frac{2}{3}=2:3

Therefore, the similarity ratio of the smaller to the larger similar cylinders is 2 : 3.

4 0
3 years ago
a cirlcular patio has a diameter of 26ft. inside it is a round flower garden with a diameter of 6ft. to edge the patio and the g
Luden [163]

Answer: C just took the test

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
Find the net pay hours worked 27 1/4 5.15 an hour
natali 33 [55]

$440.34 because 27 1/4 X 5.15 = $440.3375 rounded to nearest hundredth $440.34

3 0
3 years ago
Plz only do what i have not done
Gennadij [26K]

Answer:

7- B,A,C

8- J,A,L

9- D,E,F

10- C,D,E

11- B,A,C

12- X,Z,Y

I hope this helps!

8 0
2 years ago
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