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2 years ago

A giant tortoise can travel 0.14 miles in 1 hours .At this rate ,how long would it take to travel 2 miles ?

Mathematics

1 answer:

SCORPION-xisa [38]2 years ago

6 0

Answer:

0.28 miles

Step-by-step explanation:

Plz mark brainliest!!!

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Solve the following 2 equation. I will Make Brainliest

Levart [38]

Answer:

d. $x = 10$

e. $x=\sqrt[3]{\dfrac{15}{4}}$

Step-by-step explanation:

I've typed up my workings in MS Word and attached them (as it's very difficult to type this in the Brainly equation editor).

I've used the product, quotient and power log laws.

Product: $log_a(x)+log_a(y)=log_a(xy)$

Quotient: $log_a(x)-log_a(y)=log_a(\dfrac{x}{y})$

Power: $p\log_a(x)=log_a(x^p)$

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2 years ago

Explain why my calculator gives an error when you enter cos^-1 (5/4)

Gnoma [55]

Step-by-step explanation:

because the is no answer for that in the calculator

4 0

2 years ago

Read 2 more answers

Jordan solved the equation −7x + 25 = 48; his work is shown below. Identify the error and where it was made.

fenix001 [56]

Answer:

step 3: he should have divided both sides by -7

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2 years ago

A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only

Nadusha1986 [10]

Answer:

The sample size required is, n = 502.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

The margin of error is:

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

Assume that 50% of the people would support this political candidate.

The margin of error is, MOE = 0.05.

The critical value of z for 97.5% confidence level is:

z = 2.24

Compute the sample size as follows:

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

$n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

$=[\frac{2.24\times \sqrt{0.50(1-0.50)}}{0.05}]^{2}\\\\=501.76\\\\\approx 502$

Thus, the sample size required is, n = 502.

4 0

3 years ago

How much chromium-51 will be left after 7 months?

Makovka662 [10]

Answer:

It seems there's more information in you question that you haven't posted.

7 months = 7 * (365/12) = 213 days

The half-life of chromium 51 is 27.7 days.

After 7 months, chromium 51 will have undergone 7.69 half-lives.

To solve for the ending amount we'll use the formula:

Ending Amount = Beginning Amount / 2^n where 'n' = # of half-lives

We'll say the beginning amount =1

Ending Amount = 1 / 2^7.69

Ending Amount = 0.0048426082

Basicaly if you started with 1 gram of chromium 51, after 7 months you would have 0.0048426082 grams.

Source: https://www.1728.org/halflife.htm

Step-by-step explanation:

6 0

2 years ago