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Serga [27]
3 years ago
12

M+4-3m=4+12m alguem pode me dizer qual o resultado ?

Mathematics
1 answer:
Komok [63]3 years ago
7 0
M=0 that is what i got and I'm really good at math.
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What is the value of x?<br><br><br><br> Enter your answer in the box.
Vitek1552 [10]

Answer:

x = 27

Step-by-step explanation:

The given angles are vertical angles and congruent, thus

5(x - 4) = 4x + 7, that is

5x - 20 = 4x + 7 ( subtract 4x from both sides )

x - 20 = 7 ( add 20 to both sides )

x = 27

5 0
4 years ago
Simplify. <br> 5/√10 <br><br> A. √5 <br> B. √10/2 <br> C. √2/2 <br> D. √2
worty [1.4K]

Answer:

√10/2

Step-by-step explanation:

5/√10

Rationalizing the surd involves multiplying the numerator and denominator by the surd.

5/√10 * √10/√10

5√10/10

√10/2

8 0
3 years ago
(2.8 x 10^8) / (7 × 10^2)
raketka [301]

Answer:

(2.8 x 10^8) / (7 × 10^2) = 400'000

Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa
Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

$ 0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} } $

$ 0.2625 \pm 1.96\times 0.03478 $

$ 0.2625 \pm 0.06816 $

0.2625 - 0.06816, \: 0.2625 + 0.06816

(0.1943, \: 0.3306)

(19.43\%, \: 33.06\%)

Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

5 0
3 years ago
A milloniare estimates her wealth to be $2.2×10 to the nearst million dollars.what are the lower and upper bounds for the millio
Andrews [41]

Answer:

2.2 x10= = 22.2 =22 million.

Step-by-step explanation:

3 0
4 years ago
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