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2 years ago

The table shows the amount that each person spend on snacks, games, and souvenirs the las time they went to the carnival.

1 answer:

3 0

8/12 divide by 4 top and bottom
4/6 skipping this step
2/3 to get to here

26/x this is the end number we want

26/39 here it is multiplying the number making the number equal to the 2/3

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A store has 1173 pairs of socks. the socks are sold in packs of 4 pairs. how many packs of socks can the store sell?

Nadusha1986 [10]

Answer:

293 packs is the maximum the store can sell.

Step-by-step explanation:

Determine the number of packs that can be made with 4 pairs of socks.

(1173 pairs)/(4 pairs/pack) = 293.25 packs

We can't sell 0.25 pack (1 pair of socks), so drop that fraction to yield 293 full packs. Donate the spare pair, so to speak, to the local IRS agent.

5 0

1 year ago

Hello people who are good at math pls help me.

kherson [118]

Answer:

Q1. 12+3×2=N

12+6=N

18=N

Ans: D

Q2. 5+7×2=5+14

=19

Ans: A

Q3. 4-(3-2)+1×(15÷3)=N

4-1+1×5=N

4-1+5=N

8=N

Ans: A

Q4. 48÷(5-1)=N

48÷4=N (last step)

12=N (value of N)

Ans: B

Q5. Ans: D

6 0

2 years ago

Classify each figure on the triangle

ipn [44]

All the triangles are isosceles meaning that they have to sides that are equal

5 0

3 years ago

Read 2 more answers

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

Which trig ratio? Can you solve this??

pogonyaev

Step-by-step explanation:

There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ.

7 0

2 years ago