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Lorico [155]
3 years ago
7

(2x-1)(3x-1) in standard form

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
4 0

6x2+x−1 use the FOIL method

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Laquita practiced 60 math facts. Kinman practiced 15 more facts than Laquita. How many math facts did they practice in all?
Afina-wow [57]

Answer:

They practiced a total of 75 math facts in all!

60 + 15 = 75 :D

5 0
3 years ago
What's the answer??
Vilka [71]
The answer is neither. i hope this helps
8 0
3 years ago
2x &lt;10 or x\2 &gt;3 <br>someone knows what does x equal​
DaniilM [7]

Step-by-step explanation:

<u>For the first one</u>,

2x < 10

  • Divide both sides of the inequality by 2,

2x ÷ 2 < 10 ÷ 2

x < 5

<u>Now, for the second one</u>,

x/2 > 3

  • Multiply both sides of the inequality by 2,

x/2 × 2 > 3 × 2

this will <em>cancel</em> the <em>fraction</em> and will give us,

x > 6

That's it! I hope this explanation was useful enough ;)

8 0
3 years ago
Help ASAP! 50 points!
AURORKA [14]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 70 women ove
Yanka [14]

Answer:

Yes, evidence shows that the cream will improve the skin of more than 40% of women over the age of 50 in 0.01 significance level.

Step-by-step explanation:

We need to make a hypothesis test.

Let p be the proportion of women who used cream report skin improvement.

H_{0}: p=0.4

H_{0}: p<0.4

To test the hypothesis, we need to calculate z-score of the sample mean and compare its probability with the significance level.

z=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } } where

  • p(s) is the sample proportion of women reported improvement (0.5)
  • p is the proportion assumed under null hypothesis. (0.4)
  • N is the sample size (70)

Putting the numbers:

z=\frac{0.5-0.4}{\sqrt{\frac{0.4*0.6}{70} } } ≈ 1.71

And P(z<1.71) ≈ 0.955. Since 0.955>0.01 we fail to reject the null hypothesis that the cream will improve the skin of more than 40% of women.

8 0
3 years ago
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