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2 years ago

Please answer this question :)

Mathematics

1 answer:

tia_tia [17]2 years ago

7 0

Answer: 2.663043987
explanation: use calculator

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2 Quick algebra 1 question for 50 points!

SVETLANKA909090 [29]

Yes

Direct variation because Time increases cupcakes increases and vice versa

Constant of proportionality represents the amount of time required per cupcakes

Yes we can find it

We have to check the x and y values of the ordered pairs (x,y)

If y is decreasing with respect to increase in x then it's inverse variation

3 0

1 year ago

Write each expression in exponential form and find its value 1/6*1/6*1/6*1/6

olga55 [171]

Answer:

1/6^4.

Step-by-step explanation:

Multiply denominators.

6*6*6*6 = 1,296

Your answers are 1/6^4 and 1/1296.

3 0

3 years ago

WILL GET BRAINLEST WILL HAVE TI ZOOM IN THI SHIW WORK TOO PLSSS ASAP

Lynna [10]

Answer:

sin = 1.333333

cos = 0.75

8 0

3 years ago

A translation maps (x, y) to

Troyanec [42]

Answer:

second quadrant

Step-by-step explanation:

(x, y ) → (x - 5, y + 3 ) means subtract 3 from the original x- coordinate and add 3 to the original y- coordinate.

(- 3, - 2 ) ← in third quadrant

→ (- 3 - 5, - 2 + 3) → (- 8, 1 ) ← in second quadrant

6 0

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 - <em>α</em>)% 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 <em>z</em> 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).

5 0

3 years ago