All categories

3 years ago

What is the solution and the inequality ?

Mathematics

1 answer:

Nikolay [14]3 years ago

4 0

Inequality : 19.50n+24≤$110

Solution: Alyssa can buy 4 DVDs

Work: 110-24=86
86/19.50 = 4.4

You can't have a fraction of a DVD, so you have to round down to 4.

You might be interested in

(Correct answer only please)

andriy [413]

Answer: Choice A) Add 3.8 to both sides of the equation

Explanation:

If we knew the value of w, then we would replace it and apply PEMDAS.

However, we don't know the value of w, so we undo each step of PEMDAS going backwards.

We start with the "S" of PEMDAS, and undo the subtraction. To undo subtraction, you apply addition. To undo that "minus 3.8" we add 3.8 to both sides.

4 0

3 years ago

Read 2 more answers

Which of the following is a true statement?

sleet_krkn [62]

108/9=12 is a true statement

7 0

3 years ago

What is the unit rate of $120 15 books

Svetlanka [38]

I think its 8? someone correct me if im wrong.

3 0

3 years ago

Read 2 more answers

Helpppppo me with this pleaseeee

Troyanec [42]

Answer:

I think a but I am not 100 percent sure

Step-by-step explanation:

5 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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 weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago