F(x) = -43 + 13x

Carlotta subscribes to the HotBurn music service. She can download no more than 12 song files per week. Carlotta has already dow

UNO [17]

Answer:

Inequality: $x+10\leq 12$

Inequality solved: $x\leq 2$

The graph is attached.

Step-by-step explanation:

You need to remember the meaning of the inequalities symbols:

: Less than.

$>$ : Greater than.

$\leq$ : Less than or equal to.

$\geq$ : Greater than or equal to.

Let be "x" the number of more songs Carlotta can download.

According to the information given in the exercise, she can download no more than 12 song files per week. This indicates that you must use the symbol $\leq$ .

Since she has already downloaded 10 song files this week, you can write the following inequality that represents this situation:

$x+10\leq 12$

In order to solve it, you must subtract 10 from both sides:

$x+10-10\leq 12-10\\\\x\leq 2$

Plot this result on a number line.

Since the symbol of the inequality is $\leq$ , the dot must be filled (Observe the number line attached).

8 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider