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Eduardwww [97]
2 years ago
13

Checking account A charges a monthly service fee of $12 and a wire transfer fee of $10.50, while checking account B charges a mo

nthly service fee of $21 and a wire transfer fee of $8.50. Which check account is the better deal if four wire transfers are made per month?
Mathematics
1 answer:
son4ous [18]2 years ago
3 0

Answer:

checking account A is the better deal.

Step-by-step explanation:

A charges a monthly service fee = $12.00

Wire transfer fee = $10.50

B charges a monthly service fee = $21.00

Wire transfer fee = $8.50

If the requirement is four wire transfer per month

A charges for 4 wires = 10.50 × 4 = $42.00

and adding monthly service fees = 42.00  + 12.00 = $54.00

B charges for 4 wires = 8.50 × 4 = $34.00

and adding monthly service fees = 34.00 + 21.00 = $55.00

Therefore A charges less than B, so checking account A is the better deal.

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Kx + 3x = 4 solve for x.
victus00 [196]

Take out a common factor between 3x and kx. That means use the distributive law to get what you normally would start with.

x(k + 3) = 4

Now divide by k + 3

x = 4/(k + 3)

That's as much as you can do with this question.



4 0
2 years ago
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this surve
nasty-shy [4]

Answer:

There is enough statistical evidence to support the claim that the survey is not accurate.

Step-by-step explanation:

We have  to perform a test of hypothesis on the proportion.

The claim is that the proportion of families that own stock differs from 48.8%.

Then, the null and alternative hypothesis are:

H_0: \pi=0.488\\\\H_a:\pi\neq0.488

The significance level is 0.05.

The sample. of size n=250, has a proportion of p=0.568.

p=X/n=142/250=0.568

The standard error of the proportion is

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.488*0.512}{250}}=\sqrt{0.00099}=0.032

The z-statistic can now be calculated:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.568-0.488-0.5/250}{0.032}=\dfrac{0.078}{0.032}=2.4375

The P-value for this two-tailed test is then:

P-value=2*P(z>2.4375)=0.015

As the P-value is smaller than the significance level, the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the survey is not accurate.

6 0
2 years ago
3.82e+7 in standard form shown in calculator display
ser-zykov [4K]
E+7 means 10^+7
basically move the decimal place 7 placs to right so it would be aprox
38200000.0
3 0
2 years ago
Read 2 more answers
ANSWER ALL 5 PARTS.
N76 [4]
A function f from a set A to a set B is defined as a relation that assings to each element  x in the set A exactly one element y in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is x feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let x be the width of the tomato patch, then the function that matches this statement is:

\boxed{Wa(x)=\frac{x}{2}}

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants <span>its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

</span>\boxed{La(x)=x+2}
<span>
2. Ar</span>ea of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then t<span>he area of a rectangle is the product of the length and width. So:

</span>A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}
<span>
3. C</span><span>ombined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

</span>Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}
<span>
4. W</span>rite the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:<span>

</span>Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x

5. 
Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that <span>Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

</span>31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06
<span>
Therefore the area of the remaining space is:

</span>\boxed{Aa(7.06)=7.06ft^{2}}

6 0
2 years ago
15 PINTS!!!!!!!!! PLZ HELP!!!!!!!! The linear function f(x) = 0.5x + 80 represents the average test score in your math class, wh
Stells [14]

Answer:

Part A: The test average for the math class after completing 2 tests is 81

Part B: The test average for the science class after completing 2 tests is 83

Part C: The science class had the higher average test score after completing the test 4.

Step-by-step explanation:

The average test score for math class is given by the linear function, f(x) = 0.5·x + 80

The data for the average test score for science g(x) are;

x,   q(x)

1,     81

2,    83

3,    85

Part A: The average test score for math after completing test 2 is given as follows;

f(2) = 0.5×2 + 80 = 81

∴ The test average for the math class after completing 2 tests = 81

Part B:

The average test score for science after completing test 2 is given from the table as at x = 2, g(2) = 83

∴ The test average for the science class after completing 2 tests = 83

Part C: After completing 4 tests, we have for the math class, f(4) = 0.5×4 + 80 = 82

For the science class, it is observed that common difference between each subsequent test score average is 2, therefore, the average test score, for the fourth test is 2 added to the average test score after the third test, which gives;

Average test score, after completing the fourth test for the science class, g(4) = 85 + 2 = 87

Since g(4) > f(4) the science class had the higher average test score after completing the test 4.

4 0
2 years ago
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