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3 years ago

Find the product of 3 tens + 6 ones by 9 ones

Mathematics

1 answer:

NemiM [27]3 years ago

5 0

The answer is 324

First you must simplify the problem into an equation.

36 × 9

Then, you must multiply the 2 terms.

324

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Which of the following correctly graphs the system of equations? {y=x−3 y=−2x+3

8_murik_8 [283]

The third one

This graph is the only one that shows the correct slops with the correct y intercepts (-3 and 3)

4 0

2 years ago

In May, Jim's lunch account has a balance of $58.19. If lunch costs 2.74 per day, how many days will Jim be able to buy lunch be

velikii [3]

He will be able to buy lunch for 21 days because 58.19 divided by 2.74 is 21. 237226

5 0

3 years ago

PR and os are diameters of circle T. What is the

____ [38]

Answer:

100degrees

Step-by-step explanation:

Find the diagram attached

From the diagram

<PQT =<TRS = 40

Since the triangle STR is isosceles, here,

<TSR =<TRS = 40

Also the sum of angle in a triangle is 189, hence:

Arc SR+<TSR +<TRS = 18₩

ArcSR +40+40=180

ArcSR +80=180

ArcSR = 180-80

ArcSR = 100degrees

Hence the measure of SR is 100degrees

8 0

2 years ago

Convert 4 gallons to pints

Lena [83]

There are 8 pints in 1 gallon

In 4 gallons there are 32 pints

8 0

2 years ago

Read 2 more answers

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).

5 0

3 years ago