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Veseljchak [2.6K]
3 years ago
9

Write four and twenty-six thousandths in standard form

Mathematics
1 answer:
juin [17]3 years ago
6 0

426,000 * 10^-4. You can go use a standard form calculator. and if you don't have a physical one, you can use one online.

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154.4g of protein I'm pretty sure

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3 years ago
A tiger shark swims an average of 32 miles per hour. About how many hours will it take the tiger shark to swim 8 miles?
elena55 [62]

Answer:

0.25 hours, or 15 minutes

Step-by-step explanation:

Ok, so the rate is 32 miles per hour, so it would take about 1/4 of a hour to go 8 miles (8/32= 0.25)

6 0
3 years ago
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Molodets [167]
11:19. you would put 11 first to represent the girl and 19 by adding 11 and 8 to represent the total number of students
3 0
4 years ago
. A condo in Orange Beach, Alabama, listed for $1.4 million with 20% down and financing at 5% for 30 years. What is the monthly
vovikov84 [41]
Listed price = $1.4 million
Down payment = 20% of $1.4 million = 0.2 x 1,400,000 = 280,000
Amount left to pay = $1.4 million - 280,000 = $1,120,000
Present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / r
where: PV = $1,120,000
r = 5% = 0.05
t = 12
n = 30 years.
1,120,000 = P(1 - (1 + 0.05/12)^-(12 x 30)) / 0.05
1,120,000 x 0.05 = P(1 - (1 + 1/240)^-360)
56,000 = P(1 - 0.2238)
P = 56,000 / 0.7761 = 72,148.83
Therefore, the monthly payment is $72,148.83
8 0
4 years ago
Suppose that prices of a certain model of a new home are normally distributed with a mean of $150,000. Use the 68-95-99.7 rule t
levacccp [35]

Answer:

68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

<u><em>Let X = prices of a certain model of a new home</em></u>

SO, X ~ Normal(\mu=150,000 ,\sigma=2,300)

The z score probability distribution for normal distribution is given by;

                             Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean price = $150,000

            \sigma = standard deviation = $2,300

<u>Now, according to 68-95-99.7 rule;</u>

Around 68% of the values in a normal distribution lies between \mu-\sigma and \mu-\sigma.

Around 95% of the values occur between \mu-2\sigma and \mu+2\sigma .

Around 99.7% of the values occur between \mu-3\sigma and \mu+3\sigma.

So, firstly we will find the z scores for both the values given;

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{147,700-150,000}{2,300}  = -1

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{152,300-150,000}{2,300}  = 1

This indicates that we are in the category of between \mu-\sigma and \mu-\sigma.

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

7 0
4 years ago
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