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harkovskaia [24]
3 years ago
6

1. Alex bought 2 buckets of baseballs that cost $19.49 each, a bat that cost $125.50, and 2 pair of baseball socks that totaled

$29.50. The tax on Alex's purchases equaled $16. How much change did Alex receive if he paid with three $100 bills?​
Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

$90.02

Step-by-step explanation:

<u>Solve</u>

Formula: Change money = paid money – bill. Paid money = change + bill.

Given: <em>2</em> buckets of baseballs that cost $<em>19.49</em> each

A bat that cost $<em>125.50</em>, and <em>2</em> pairs of baseball socks that totaled $<em>29.50</em>

Tax on Alex's purchases equaled $<em>16</em>.

Paid with three $<em>100</em> bills.

To find: How much change Alex received if he paid with three $100 bills

Total payment + Total Tax  -Total cost

19.49 + 125.50 + 29.50  = 174.49 (Don't add the $16 tax because that will be taken from the amount we pay)

∴, <em>three </em>100 bills is 300

So, 300 - 174.49 = 109.51

109.51 - 16 = 90.51

90.51 - 0.49 = $90.02

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A cone-shaped paper drinking cup is to be made to hold 27 cm of water. Find the height h and radius r of the cup that will use t
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Step-by-step explanation:

The formula for the volume of a cone of radius r and height h is ...

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Then r² can be found in terms of h and V as ...

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This will be zero when ...

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The optimal radius is 2.632 cm; the optimal height is 3.722 cm.

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Read 2 more answers
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

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(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

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(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

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                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

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