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Lorico [155]
3 years ago
12

In March, Isaiah and Freddie had the same amount of money in their savings accounts. In April, Isaiah deposited $210 into his ac

count. Freddie increased the money in his account by 15%. When they compared their balances, they found that they were still equal.
How much money did they both have in their accounts in March?
Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

They both have in their accounts in March $1400

Step-by-step explanation:

They had the same amount of money in their savings in March.Next month, Isaiah deposited $210. At the same time, Freddie increased the money by 15%.

After that, they compared again, and they found that they were still equal. So, we can conclude that 15% of the initial amount is $210 because the money was the same in March and April.Now, we need to calculate 100%. We know 15% to be $210.

210 * (100/15) = 1400.

Finally, we can conclude that the initial amount for both corresponds to 100% and it is $1400. 

Artemon [7]3 years ago
7 0
If they were equal both times they checked this means that 15% of their $ was equal to $210
so 15/100   = 210/x
so .15x=210
x=210/.15 
x = 1400

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4 years ago
Write the phrase as an expression. Then evaluate when x = 3 and y = 15
victus00 [196]

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Put x then y u get x+y=15*2

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3 years ago
Maya bought a new iphone case for $40. Now, it is on sale for $25. Find the percent of change in the
bulgar [2K]

Answer:

Step 1: $15

Step 2: 38%

Step-by-step explanation:

Step 1:

40 - 25 =15.

Therefore, the price would decrease by $15.

Step 2:

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4 0
3 years ago
3. Suppose the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 custo
stich3 [128]

Answer:

The probability that the average number of customers in the sample is between 290 and 310 is 0.61922.

Step-by-step explanation:

We are given that the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 customers.

A random sample of 50 business days was selected.

<em>Let </em>\bar X<em> = sample average number of customers</em>

The z-score probability distribution for sample average is given by;

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

where, \mu = population average = 301 customers

            \sigma = population standard deviation = 80 customers

            n = sample of business days = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

(a) Probability the average number of customers in the sample is between 290 and 310 is given by = P(290 < \bar X < 310) = P(\bar X < 310) - P(\bar X \leq 290)

   P(\bar X < 310) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{310-301}{\frac{80}{\sqrt{50} } } ) = P(Z < 0.79) = 0.78524

   P(\bar X \leq 290) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{290-301}{\frac{80}{\sqrt{50} } } ) = P(Z \leq -0.97) = 1 - P(Z < 0.97)

                                                       = 1 - 0.83398 = 0.16602                       

<em>{Now, in the z table the P(Z  </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.79 and x = 0.97 in the z table which has an area of 0.78524 and 0.83398 respectively.}</em>

Therefore, P(290 < \bar X < 310) = 0.78524 - 0.16602 = 0.61922

<u><em>Hence, the probability that the average number of customers in the sample is between 290 and 310 is 0.61922.</em></u>

4 0
3 years ago
Can anyone help me with this question​
Damm [24]

The domain of a function is all the possible input or x values.

In this problem, we can see the function starts at (0, 0) tapering off to the right infinitely. This means that every positive x value including 0 is in the domain.

0, 1, 2, 5 are all in the domain.

8 0
4 years ago
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