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Salsk061 [2.6K]

3 years ago

7

the cost C (in dollars) for beads to make bracelets is C = 4x + 180, where x is the number of bracelets. Each bracelet sells for

$34. Write an equation for the revenue R in terms of the number of bracelets

1 answer:

luda_lava [24]3 years ago

3 0

R = 34x - (4x + 180)

If you liked my answer please make me the brainliest answer! :)

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greta has a piece of cloth that is 9 yards long . she cuts it into pieces that are each 1/3 yards long how many pieces of cloth

labwork [276]

27 peices i believe. 3 peices per yard, 9 yards. 9*3=27

7 0

1 year ago

The depreciating value of a semi-truck can be modeled by y = Ao(0.84)x, where y is the remaining value of the semi, x is the tim

NemiM [27]

The value of the truck initially, Ao is
83000

1-0.16=0.84
1-0.26=0.74

After one year the value
Y=83,000×(0.84)=69,720
Y=83,000×(0.74)=61,420
When you compare the results you will see that the graph would fall at a faster rate to the right because the depreciation rate of 26% is higher than the depreciation rate of 16%

Hope it helps

8 0

4 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, <u><em>the test statistics</em></u> = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago

Simply 7/4 squared

klemol [59]

Answer:

0.4375

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Let B1W2 denote the outcome that the first ball drawn is B1 and the second ball drawn is W2. Because the first ball is replaced

Sati [7]

Answer:

$\{B_1B_1, B_1B_2, B_1W_1, B_1W_2, B_1W_3, \\B_2B_1, B_2B_2, B_2W_1, B_2W_2, B_2W_3, \\$

$W_1B_1,W_1B_2, W_1W_1, W_1W_2, W_1W_3, \\W_2B_1, W_2B_2, W_2W_1,W_2W_2, W_2W_3, \\W_3B_1, W_3B_2, W_3W_1, W_3W_2, W_3W_3\}$

Step-by-step explanation:

$\text{If an urn contains two blue balls} (denoted \:B_1 \:and \:B_2) \text{and three white balls}$ , $(denoted \:W_1, W_2, \:and\: W_3)$

If One ball is drawn, its color is recorded, and it is replaced in the urn. Then another ball is drawn and its color is recorded.

The 25 Possible outcomes of this experiment are listed below:

$\{B_1B_1, B_1B_2, B_1W_1, B_1W_2, B_1W_3, \\B_2B_1, B_2B_2, B_2W_1, B_2W_2, B_2W_3, \\$

$W_1B_1,W_1B_2, W_1W_1, W_1W_2, W_1W_3, \\W_2B_1, W_2B_2, W_2W_1,W_2W_2, W_2W_3, \\W_3B_1, W_3B_2, W_3W_1, W_3W_2, W_3W_3\}$

The tree diagram of this event is also attached.

6 0

3 years ago