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2 years ago

Bill is ten years older than his sister. If Bill was twenty-five years of age in 1983, in what year could he have been born?

Mathematics

2 answers:

Andre45 [30]2 years ago

3 0

Answer:

1968

Step-by-step explanation:

Elis [28]2 years ago

3 0

Heyy,

83-25=58
Bill would of been born in 1958

Hope this helps :)

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Please helpppp! BRAINLIEST to correct answer!!!

OLga [1]

Instead of negative positive and instead of positive negative hope you get it right. Am not that smart

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3 years ago

By the age of 21, the best violinist and pianist will have practiced at least 10,000 hours. If you practice and instrument 45 mi

Alika [10]

Answer:

273.94 hours.

Step-by-step explanation:

1 day = 0.75 hours of practicing in a day.

365.25 = x

1/365.25 = 0.75/x Cross multiply

x = 0.75 * 365.25 Combine

x = 273.94 hours

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3 years ago

When is interest best compounded?

e-lub [12.9K]

Usually, the more often it's compounded, the better. That's assuming that the percent is equal through all comparisons, however.

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2 years ago

We measure the diameters for a random sample of 25 oak trees in a neighbourhood. Diameters of oak trees in the neighbourhood fol

jarptica [38.1K]

Answer:

The required confidence inteval = 94.9%.

Step-by-step explanation:

Confidence interval: Mean ± Margin of error

Given: A confidence interval for the true mean diameter of all oak trees in the neighbourhood is calculated to be (36.191, 42.969).

i.e. Mean + Margin of error = 42.969 (i)

Mean - Margin of error = 36.191 (ii)

Adding (i) and (ii), we get

$2Mean =79.16\\\\\Rightarrow\ Mean= 39.58$

Margin of error = 42.969-39.58 [from (i)]

= 3.389

Margin of error = $t^* \dfrac{\sigma}{\sqrt{n}}$

here n= 25 $, \ \sigma=8.25$

i.e.

$3.389=t^*\dfrac{8.25}{5}\\\\\Rightarrow\ t^* = \dfrac{3.389}{1.65}\\\\\Rightarrow\ t^* =2.0539 \$

Using excel function 1-TDIST.2T(2.054,24)

The required confidence inteval = 94.9%.

7 0

2 years ago

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2885 and standard deviation

aliina [53]

Answer:

a. X~N(2,885, 651)

b. 0.086291

c. 0.00058

d. 3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

$P(X$

#substitute the given values in the formula to solve for P:

$P(X$

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

$P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058$

Hence, the proportion of customers consuming over 5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

$0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09$

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

8 0

3 years ago