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3 years ago

(6×1/10)+(3×1/100)+(9×1/1000)

Mathematics

1 answer:

NISA [10]3 years ago

5 0

Answer:

0.639

Step-by-step explanation:

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Seven plus five times a number is gteater than or equal to negative 9

zubka84 [21]

Step-by-step explanation:

7+5(x) ( between here write the greater than or equal to sign) -9

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3 years ago

Solve for x A 130 B 120 C 23 D 70

34kurt

Answer:

B)120º

Step-by-step explanation:

70+50=x

120=x

7 0

2 years ago

Help pls, will choose brainliest if explanation is provided

Nookie1986 [14]

Answer:

$solve \: when \: (f + g) \:(4) \\ \\ first \: solve \: for \: f(x) \\ \: \\ solution \: .... \\ \\ f(x) = 4x - 3 \\ f(4) = 4(4) - 3 \\ f(4) = 16 - 3 \\ f(4) = 13 \\ \\ \\ now \: solve \: for \: g(x) \\ \: \\ solution \: .... \\ \\ g(x) = {x}^{3} + 2x \\ g(x) = {(4)}^{3} + 2(4) \\ g(x) = 64 + 8 \\ g(x) = 72 \\ \\ (f) - (g) = 13 - 72 \\ (f) -(g) = -59$

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2 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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3 years ago

Describe the transformation from f(x) = x⁴ to g(x) = (x - 2)⁴ - 3

pogonyaev

Answer:

Two to the left, three down

Step-by-step explanation:

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3 0

2 years ago