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nata0808 [166]
1 year ago
15

Using a rounded version of a conversion ratio allows for a quick estimate of a unit conversion

Mathematics
1 answer:
dolphi86 [110]1 year ago
6 0

The conversion ratio that can be illustrated will be converting 2030m to kilometers.

<h3>How to illustrate the conversion?</h3>

The unit conversion that can be illustrated based on the information can be convert 2030m to kilometers.

It should be noted that 1000 meters makes 1 kilometer. Therefore, rounding up 2030m mentally will be 2000m and the conversion to kilometers will be:

= 2000/1000

= 2 km

Learn more about ratio on:

brainly.com/question/2328454

#SPJ1

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Paul needs 40 pounds of 48% silver alloy to finish a collection of jewelry. How many pounds each of 45% silver alloy and 60% sil
Salsk061 [2.6K]
<span>Equation:
silver + silver = silver
0.45x + 0.60(40-x) = 0.48*40
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45x + 60*40 - 60x = 48*40

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-15x = -12*40
x = (4/5)40
x = 32 lbs (amt of 45% slloy needed)
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8 0
2 years ago
Please help on question b
Viktor [21]

Answer:

Step-by-step explanation:

Number of males who exclusively drink coffee: 21.

Total number of people: 100

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8 0
3 years ago
Solve for a, v0, and t
Burka [1]

Answer:

Step-by-step explanation:

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5 0
3 years ago
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, <em>MOE</em> = 0.04.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0
3 years ago
Directions: Select all the correct answers.
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The correct answer is C and D .
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