The original cost of an item is $64, but you have to pay $78.08. What is the markup of the item (as a percent)?

Chris has a large collection of hockey cards and wants to get rid of some of his hockey cards he gives 2 cars away on day 1,4 ca

Marizza181 [45]

Answer:

I'm not sure for which answer you're asking for so I'll give you all of them. On the tenth day, Chris will give away 1024 cards.

Step-by-step explanation:

On the fourth day he will give away 16, on the fifth day 32, on the sixth day 64, on the seventh say 132, on the eighth day 264, on the ninth day 528, and on the tenth day 1024

6 0

2 years ago

Choose the function whose graph is given by:

loris [4]

<h2>Answer:</h2>

$y=3sec\left(\frac{1}{2}x\right)$

<h2>Step-by-step explanation:</h2>

The graphs of $sec(x)$ can be obtained from the graph of the cosine function using the reciprocal identity, so:

$sec(x)=\frac{1}{cos(x)}$

But in this problem, the graph stands for the function:

$y=3sec\left(\frac{1}{2}x\right)$

Because the period is now 4π as indicated and for $x=0$ in the figure and this can be proven as follows:

$Period=\frac{2\pi}{\frac{1}{2}}=4\pi$

Also, $for \ x=1 \ then \ y=3$ as indicated in the figure and this can be proven as:

$y=3sec\left(\frac{1}{2}x\right) \\ \\ y=\frac{3}{cos(0.5x)} \\ \\ y=\frac{3}{cos(0.5(0))} \\ \\ y=\frac{3}{1}=3$

5 0

2 years ago

Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ of the

Marizza181 [45]

Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

6 0

2 years ago

Segment GI is congruent to Segment JL and Segment GH is congruent to Segment KL. I have to prove Segment HI is congruent to Segm

madreJ [45]

Answer:

See explanation

Step-by-step explanation:

1 step: $\overline{GI}\cong \overline {JL}$ - given

2 step: $\overline{GI}\cong \overline{GH}+\overline{HI}$ - Segments Addition Postulate

3 step: $\overline{GH}+\overline{HI}\cong \overline {JL}$ - Substitution Property

4 step: $\overline {JL}\cong \overline {JK}+\overline {KL}$ - Segments Addition Postulate

5 step: $\overline{GH}+\overline{HI}\cong \overline {JK}+\overline {KL}$ - Substitution Property

6 step: $\overline{GH}\cong \overline {KL}$ - given

7 step: $\overline{GH}+\overline{HI}\cong \overline {JK}+\overline {GH}$ - Substitution Property of Equality

8 step: $\overline{HI}\cong \overline {JK}$ - Subtraction Property of Equality

3 0

2 years ago

A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If

dsp73

Answer: B) 4.963±0.019.

Step-by-step explanation:

Confidence interval for population mean ( when population standard deviation is not given) is given by :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

$\overline{x}= 4.963\ lb$

s= 0.067 lb

For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.

Now , substitute all values in the formula , we get

$4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019$

Hence, a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is $4.963\pm0.019$ .

Thus , the correct answer is B) 4.963±0.019.

7 0

2 years ago