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tino4ka555 [31]
2 years ago
11

Is it - or + 60 plz help me

Mathematics
1 answer:
Artemon [7]2 years ago
6 0
+60 two negatives equal a positive
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Solve for x.
olga nikolaevna [1]

Answer:

\boxed{x=5},x=-9

Step-by-step explanation:

To rid the denominators, multiply both sides by (x-3)(x+1).

We get:

6(x+1)-12(x-3)=(x-3)(x+1)

Simplifying, we have:

6x+6-12x+36=x^2-3x+x-3,\\-6x+42=x^2-2x-3,\\x^2+4x-45=0

Solving, we get:

x^2+4x-45=0,\\(x-5)(x+9)=0,\\x=5,x=-9

Therefore, the solution with the greatest value is x=\boxed{5}

8 0
2 years ago
2 1/2 * 1 1/6=???<br>I need help can someone explain this for me?
Dovator [93]

Answer:

35/12

Step-by-step explanation:

2 1/2 is 5/2

1 1/6 is 7/6

Multiply these together to get (5*7)/(2*6) or 35/12.

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
How would i go about simplifying (x+h)^3?
Sindrei [870]
H^3+3h^2x+3hx^2+x^3 should be the answer
7 0
2 years ago
Mary is writing invitations for her birthday party. She plans to invite 16 friends. So far, Mary has written 12 invitations. Whi
love history [14]

Answer:

The answer is 75% and 75/100

8 0
3 years ago
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