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gogolik [260]
3 years ago
15

How to equivalent 64 1/2​

Mathematics
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

129/2

Step-by-step explanation:

      1                129

64  ---     =       -------

      2                 2

To find the improper fraction, you multiply the denominator by the whole number and then add the numerator.  In the end, you put that answer over the starting denominator.

64 X 2 = 128      128 + 1 = 129     129 / 2 is the answer

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MA_775_DIABLO [31]
Multiply it by its self 3 times for example 2 to the square root of 3 2x2x2
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What is this awnser? 2/5y=4
MA_775_DIABLO [31]

Answer:

\frac{2}{5}y=4\quad :\quad y=10

4 0
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Read 2 more answers
Find the distance between the points ( – 9, – 4) and ( – 4,8).
Dmitriy789 [7]

Answer:

Distance=13\ units

Step-by-step explanation:

<u><em>Distance between two points:</em></u> Distance between two pints (x_1,y_1)\ and\ (x_2,y_2) is given by

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between (-9,-4)\ and\ (-4,8).

d=\sqrt{(-4-(-9))^2+(8-(-4))^2}\\\\d=\sqrt{(-4+9)^2+(8+4)^2}\\\\d=\sqrt{(5)^2+(12)^2}=\sqrt{25+144}\\\\d=\sqrt{169}\\\\d=13

6 0
3 years ago
• 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
What is the value of x in the following equation?<br> 2 (x - 4) + 15 = 21
viktelen [127]
The answer would be x=7.
5 0
3 years ago
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