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2 years ago

How many years does Samsung galaxy 2 in 1 laptop last

Mathematics

1 answer:

patriot [66]2 years ago

7 0

Harging battery that lasts 16.5 hours¹ and innovations galore, this Galaxy Book was made to exceed all expectations.

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Solve this system of equations

olga_2 [115]

Answer:

D) x = 4, y = -2, z = 3

Step-by-step explanation:

x = 3z − 5

2x + 2z = y + 16

2(3z - 5) + 2z = y + 16

6z - 10 + 2z = y + 16

8z = y + 26 ---> (A)

7x − 5z = 3y + 19

7(3z - 5) - 5z = 3y + 19

21z - 35 - 5z = 3y + 19

16z = 3y + 54 ---> (B)

8z = y + 26

16z = 3y + 54

2(y + 26) = 3y + 54

2y + 52 = 3y + 54

y = -2

8z = -2 + 26

8z = 24

z = 3

x = 3(3) - 5

x = 4

8 0

3 years ago

Is 5.3 repeating rational

Goshia [24]

Repeating decimals are considered rational numbers because they can be represented as a ratio of two integers. If a number is terminating or repeating, it must be rational if a decimal is both non terminating and non repeating, the number is irrational.

So yes.

4 0

4 years ago

Round to the nearest 10.<br> 473 <br>91<br>889<br> 124

Novosadov [1.4K]

470

890

120

hope this helped :)

please mark as brainliest !!

3 0

3 years ago

Read 2 more answers

Brainliest ,easy question

insens350 [35]

Answer:

Scale factor is 2/5 or 0.4

Step-by-step explanation:

Divide radius of smaller circle (2) by radius of large circle (5)

4 0

3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago