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jek_recluse [69]

4 years ago

14

15. Determine whether the solution space of the system Ax = 0 is a line through the origin, a plane through the origin, or the o

rigin only. If it is a plane, find an equation for it. If it is a line, find parametric equations for it. (a) A = ⎡⎢⎣ −1 1 1 3 −1 0 2 −4 −5 ⎤⎥⎦ (b) A = ⎡⎢⎣ 1 2 3 2 5 3 1 0 8 ⎤⎥⎦

1 answer:

ladessa [460]4 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Construct a 95% confidence interval for the population mean, mu. Assume the population has a normal distribution. A random sam

aniked [119]

Given Information:

Number of lithium batteries = n = 16

Mean life of lithium batteries = μ = 645 hours

Standard deviation of lithium batteries = σ = 31 hours

Confidence level = 95%

Required Information:

Confidence Interval = ?

Answer:

$CI = 628.5 \: to \: 661.5 \: hours$

Step-by-step explanation:

The confidence interval is given by

$CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})$

Where μ is the mean life of lithium batteries, σ is the standard deviation, n is number of lithium batteries selected, and t is the critical value from the t-table with significance level of

tα/2 = (1 - 0.95) = 0.05/2 = 0.025

and the degree of freedom is

DoF = n - 1 = 16 - 1 = 15

The critical value (tα/2) at 15 DoF is equal to 2.131 (from the t-table)

$CI = 645 \pm 2.131(\frac{31}{\sqrt{16}})$

$CI = 645 \pm 2.131(7.75})$

$CI = 645 \pm 16.515$

$CI = 645 - 16.515 \: and \: 645 + 16.515$

$CI = 628.5 \: to \: 661.5 \: hours$

Therefore, the 95% confidence interval is 628.5 to 661.5 hours

What does it mean?

It means that we are 95% confident that the mean life of 16 lithium batteries is within the interval of (628.5 to 661.5 hours)

6 0

3 years ago

Simplifying fractions Sheet 2<br>

raketka [301]

<u>Simplifying fractions:</u>

1) $\frac{14}{20} =\frac{7}{10}$ 2) $\frac{4}{8} =\frac{1}{2}$

3) $\frac{9}{12} =\frac{3}{4}$ 4) $\frac{12}{15} =\frac{4}{5}$

5) $\frac{8}{18} =\frac{4}{9}$ 6) $\frac{14}{21} =\frac{2}{3}$

7) $\frac{12}{16} =\frac{3}{4}$ 8) $\frac{10}{24} =\frac{5}{12}$

9) $\frac{15}{35}=\frac{3}{7}$ 10) $\frac{13}{26} =\frac{1}{2}$

11) $\frac{11}{55}=\frac{1}{5}$ 12) $\frac{9}{21} =\frac{3}{7}$

13) $\frac{16}{26} =\frac{8}{13}$ 14) $\frac{20}{32} =\frac{5}{8}$

15) $\frac{18}{24} =\frac{3}{4}$ 16) $\frac{21}{27} =\frac{7}{9}$

17) $\frac{4}{32} =\frac{1}{8}$ 18) $\frac{25}{40} =\frac{5}{8}$

<em>Hope this helps :)</em>

5 0

3 years ago

-27-(-8)=<br><br>Thank you if u answer

Anastaziya [24]

-27+8
( The minus signs multiply each other to form a +. )
= -19

7 0

4 years ago

Read 2 more answers

A wholesaler requires a minimum of 4 items in each order from its retail customers. The manager of one retail store is consideri

Leto [7]

Answer:

The 1st graph will be perfectly matching the conditions.

Step-by-step explanation:

A wholesaler requires a minimum of 4 items in each order from its retail customers.

The manager of one retail store is considering ordering a certain number of sofas, x and a certain number of pillows that come in pairs, y.

Therefore, for y = 0, x ≥ 4 and for x = 0, y ≥ 2 to make the order of 4 items or more.

Therefore, the equation for the order to be of 4 items will be

x + 2y = 4.

So, the 1st graph will be perfectly matching the conditions. (Answer)

3 0

3 years ago

Suppose that 5 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 39 cm.

vesna_86 [32]

Answer

given,

work = 5 J

spring stretch form 30 cm to 39 cm

$W = \dfrac{1}{2}kx^2$

x = 0.39 - 0.30 = 0.09 m

$5 = \dfrac{1}{2}k\times 0.09^2$

$k = \dfrac{5\times 2}{0.09^2}$

k = 1234.568 N/m

a) work when spring is stretched from 32 cm to 34 cm

x₂= 0.34 -0.30 = 0.04 m

x₁ = 0.32 - 0.30 = 0.02

$W = \dfrac{1}{2}k(x_2^2-x_1^2)^2$

$W = \dfrac{1}{2}\times 1234.568 \times (0.04^2-0.02^2)^2$

W = 0.741 J

b) F = k x

25 = 1234.568 × x

x = 0.0205 m

x = 2.05 cm

3 0

4 years ago