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3 years ago

6

Two small pitchers and one large pitcher can hold 8 cups of water. One large pitcher minus one small

1 answer:

Kitty [74]3 years ago

7 0

Answer:

20 cups

Step-by-step explanation:

Let the small pitchers be given by s

and the larger pitchers by l

According to the given information:

$2s + l = 8....(1) \\ and \\ l - s = 2 \\ l = s + 2...(2) \\ substituting \: l = s + 2 \: in \: equation \: (1) \\ 2s + s + 2 = 8 \\ 3s = 8 - 2 \\ 3s = 6 \\ s = \frac{6}{3} \\ \huge \red{ s = 2} \\ substituting \: s = 2 \: in \: equation \: (2) \\ l = 2 + 2 \\ \huge \purple{ l = 4}\\\\Now\\ 3l +4s \\= 3\times 4+4\times 2\\= 12 +8\\= 20\: cups$

So, 3 large pitchers and 4 small pitchers hold 20 cups of water.

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The cafeteria has 6 round tables and 23 rectangular chairs tables. If each round table has 7 chairs ,and each rectangular table

shtirl [24]

Answer:

456

Step-by-step explanation:

Number of chairs around the round tables = 6×7=42

Number of chairs around the rectangular tables =23×18=414

The total number of chairs in the cafeteria = 6×7+23×18=456

4 0

3 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

$SE_{p_A-p_B} = 0.050$

5 0

2 years ago

How would I do this question.

olga2289 [7]

Answer:

The density of cube is 6.17959 $\frac{g}{cm^{3} }$

Step-by-step explanation:

The density is given by ration of a mass of body and volume occupied by a body.

$\rho = \frac{m}{V}$

Where,

$\rho$ is density.

m is mass of a body

V is the volume of a body

Given that one side of the cube is 0.53cm

Hence, the volume of a body is V= $0.53^{3}$ =0.148877 $cm^{3}$

Now, the Density of the cube will be

$\rho = \frac{m}{V}$

$\rho = \frac{0.92}{0.148877}$

$\rho =6.17959$

Thus, The density of cube is 6.17959 $\frac{g}{cm^{3} }$

3 0

3 years ago

A submarine dives 320 feet below the water's surface. Over the next six

sergeinik [125]

Answer:

80

Explanation:

Since it rises at a rate of 40 feet per minute, just multiply 40 times 6 to get 240. Then subtract 320 minus 240 to get 80

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2 years ago

How many questions will the equation 5(3x-9)=(6x+9) have

Delicious77 [7]

What do you mean?????????????

8 0

2 years ago