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3 years ago

On average a herd of elephants travels 10 miles in 12 hours(picture included)

Mathematics

2 answers:

Artyom0805 [142]3 years ago

6 0

Answer:

The answer is in this picture you will thank me later :)

Step-by-step explanation:

Tema [17]3 years ago

4 0

For the first Q the answer is the first square on the left 10 miles divided by 12 hours = 0,83 miles

And the second Q goes with the last square 12 hours divided by 10 miles = 1.2 h which in minutes is 72 minutes or 1h and 12 minutes.

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Create a budget that will allow you to save at least $100 by the end of October.

insens350 [35]

According to the updated budget, the net income should be at least $2100.

A budget is an estimated strategy for a certain timeframe, mostly one year. Additionally, projected material amounts, expenditures, liabilities, holdings, responsibilities, and working capital may be included.

The following list explains why the figures shown above are accurate:

The factors for the costs in the previous budget are;

Net income =$1,850.00

Cost of rent =$600

Gas, vehicle payment, and insurance together = $475.

Cost of phone, online world, and utility companies = $230.

Cost of grocery stores= $300

Cost of amusement= $50

Cost of homeowners insurance = $20.

$75 has been set aside for recreational expenditures.

Net monthly savings = $100.

Modifications to the spending plan:

$50 is the allotted sum for a computer system.

Monthly net savings = $300

∴Savings increases= $300-$100=$200

Previous net income+ cost of computer system+ 200(extra savings than previous)

= $(1850+50+200)

=$2100

So,Net earnings = $2,100

Learn more about the budget plan:

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3 0

2 years ago

The inequalities x + 3 < -5 and -x > -8 are the same.<br><br> True False

trapecia [35]

Answer:

False

Step-by-step explanation:

x + 3 < -5

x < -8

-x > -8

x < 8

Both aren't same

3 0

2 years ago

Two accountants for the firm of Elwes and Wright are arguing about the merits of presenting an income statement in a multiple-st

olchik [2.2K]

Answer:

73,188

Step-by-step explanation:

fees plus costs minus expenses

8 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{X - 3.6}{0.4}$