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2 years ago

Samuel is 4 1/6 feet tall. His dad is 5 3/4 feet tall.How much taller is Samuels dad then Samuel

Mathematics

1 answer:

Sloan [31]2 years ago

6 0

Answer:

1 7/12

Step-by-step explanation:

Just subtract the fractions if you cant do it by head or paper use a calc so 5 3/4 - 4 1/6= 1 7/12 samuels dad is 1 7/12 taller than samuel.

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Kay [80]

Answer:

C) 84

Step-by-step explanation:

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2 years ago

At Sara's new job she spent $11.52, $6.48, $5.99, $14.00, and $9.50 on lunch the first week. In the second week, she spent $4 mo

elena55 [62]

The total amount that Sara spent for her meals during the first week is,
$11.52 + $6.48 + $5.99 + $14.00 + $9.50 = $47.49
With an average of $9.498

During the second week, she spent $4 more making her total equal to $51.49. The average is equal to $10.298. The increase in the mean is equal to,
increase in mean = $10.298 - $9.498 = $0.8<span>Comments</span>

4 0

2 years ago

Read 2 more answers

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

6 0

2 years ago

The body temperatures of adults have a mean of 98.6°F and a standard deviation of 0.60° F. Describe the center and variability o

amid [387]

Answer:

center = 98.6, variability = 0.08

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

The center is the mean.

So $\mu = 98.6$

The standard deviation of the sample of 50 adults is the variability, so

$s = \frac{0.6}{\sqrt{50}} = 0.08$

So the correct answer is:

center = 98.6, variability = 0.08

7 0

3 years ago

You have been averaging 55 sales per day. Per our new policy, everyone needs to increase his or her sales per day by 10% in the

Mariulka [41]

To find your new number of sales per month, you will need to find the number of sales you have per month (based on 30 days per month).

55 x 20 = 1650 sales.

Than, you will need to calculate what 10% of this is and add it to your original sales number.

0.1 (10%) x 1650 = 165

1650 + 165 = 1815

You will need approximately 1815 sales based on the 30 day month.

8 0

2 years ago