Can someone please help me with this?

Julli [10]

Answer:

3x - 4 and 2x + 15

Step-by-step explanation:

7 0

2 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

2 years ago

2. Find the gross monthly income from this job and calculate your monthly take-home pay.

Step2247 [10]

Answer:

you will need to know what's his gross monthly income and multiply by 25%that will give you the amount of taxes withheld and then subtract that from gross monthly income and that will give you Monthly take home pay.

4 0

3 years ago

6) The amount of Jen's monthly phone bill is normally distributed with a mean of $70 and a standard

alexandr402 [8]

Answer:

i)The correct option is D.) 99.74%

ii) The correct option is B.) 68.26%

Step-by-step explanation:

i) P(10≤X≤70) = P( (10−40)/10 ≤Z≤ (70−40 )/10 ) = Pr(−3≤Z≤3)

= 0.9987 - 0.0013 = 0.99734

Therefore the percentage of Jen's monthly phone bills are between $40 and $100 is D.) 99.74%

ii)P(2.1≤X≤3.1) = P( (2.1 − 2.6) /0.5 ≤ Z ≤ (3.1−2.6 )/0.5) = Pr(−1 ≤Z ≤1) )

= 0.8413 − 0.1587 = 0.6826

Therefore the percentage of students at college have a GPA between 2.1 a,d 3.1 is B.) 68.26%

5 0

2 years ago

Why is (x-3) NOT a factor of x3+2x2-5x-6

ivolga24 [154]

As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Step-by-step explanation:

Given expression is:

$x^3+2x^2-5x-6$

We have to check if x-3 is a factor of given expression

In order for x-3 to be a factor of expression, we have to put x-3 = 0 => x=3 in expression.

If the expression is zero at x=3 then x-3 is a factor of expression otherwise not.

So putting x=333 in expression

$=(3)^3+2(3)^2-5(3)-6\\= 27+2(9) - 15-6\\= 27+18-15-6\\= 45-21\\=24 \neq 0$

As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Keywords: Polynomials, expressions

Learn more about polynomials at:

#LearnwithBrainly

8 0

3 years ago