The function f(x) = |x| is graphed over the interval [−6, 3].

Florence is itemizing deductions on her federal income tax return. Her AGI was $197,130 last year, and she contributed $99,310 t

zhannawk [14.2K]

Adjusted gross income (AGI) is a person's total gross income minus specific deductions. So, Florence can deduct an amount of: .5 or 50% of 197130 = 98,565 is the maximum deduction which is less that actual amount given for the charitable contributions.

6 0

4 years ago

Mike can build a go cart in 30 days, Jed can build one in 20 days and Erick in 60 days. If all of them share the work, how quick

Gwar [14]

Answer: 10 days

Step-by-step explanation:

Mike: $\frac{x}{30}$

Jed: $\frac{x}{20}$

Erick: $\frac{x}{60}$

Together: $\frac{x}{30}$ + $\frac{x}{20}$ + $\frac{x}{60}$ = 1

60 $(\frac{x}{30}$ + $\frac{x}{20}$ + $\frac{x}{60}$ = 1)

2x + 3x + x = 60

6x = 60

x = 10

4 0

3 years ago

Read 2 more answers

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

solniwko [45]

For this case we have to:

Let $u = x + 1$

So:

$u ^ 2-4u + 2 = 0$

We have the solution will be given by:

$u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 1\\b = -4\\c = 2$

Substituting:

$u = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (1) (2)}} {2 (1)}\\u = \frac {4 \pm \sqrt {16-8}} {2}\\u = \frac {4 \pm \sqrt {8}} {2}\\u = \frac {4 \pm \sqrt {2 ^ 2 * 2}} {2}\\u = \frac {4 \pm2 \sqrt {2}} {2}$

The solutions are:

$u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}$

Returning the change:

$2+ \sqrt {2} = x_ {1} +1\\x_ {1} = 1 + \sqrt {2}\\2- \sqrt {2} = x_ {2} +1\\x_ {2} = 1- \sqrt {2}$

Answer:

$x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}$

3 0

3 years ago

Read 2 more answers

What is the simplified value of the exponential expression 27 1/3? 1/3 1/9 3 9

madreJ [45]

Answer:

Step-by-step explanation:

27^1/3

27 is cube root of 3

so it can also be written as (3)^3

∴ {(3)^3}^1/3

3 and 3 will get cancelled

so it will be 3^1

= 3

8 0

4 years ago

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales

mihalych1998 [28]

Answer:

We conclude that the mean number of calls per salesperson per week is more than 37.

Step-by-step explanation:

We are given that the Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 37 sales calls per week on professors.

To investigate, a random sample of 41 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 5.6 calls.

Let $\mu$ = true mean number of calls per salesperson per week.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 37 {means that the mean number of calls per salesperson per week is less than or equal to 37}

Alternate Hypothesis, $H_A$ : $\mu$ > 37 {means that the mean number of calls per salesperson per week is more than 37}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of calls made last week = 40

s = sample standard deviation = 5.6 calls

n = sample of sale representatives = 41

So, test statistics = $\frac{40-37}{\frac{5.6}{\sqrt{41} } }$ ~ $t_4_0$

= 3.43

Hence, the value of test statistics is 3.43.

Now at 0.025 significance level, the t table gives critical value of 2.021 at 40 degree of freedom for right-tailed test. Since our test statistics is more than the critical value of t as 3.43 > 2.021, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the mean number of calls per salesperson per week is more than 37.

4 0

3 years ago