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2 years ago

I need an answer! NO LINKS PLS! il give brainlist

Mathematics

1 answer:

cricket20 [7]2 years ago

4 0

Answer:

D. The company's chocolate bars weigh 3.2 ounces on average.

Step-by-step explanation:

We are given that a company claims that its chocolate bars weigh 3.2 ounces on average.

The company took many large samples, and each time the mean weight of the sample was within the 95% confidence interval.

Definition of 95% confidence level: 95% confidence level means a range of values that you can be 95% certain contains the true mean of the population.

Thus by considering definition we can conclude that The company's chocolate bars weigh 3.2 ounces on average.

Thus Option D is correct.

D. The company's chocolate bars weigh 3.2 ounces on average.

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Dustin loves to cook with fresh herbs. So, he decides to plant 8 different herbs and keep the small pots on his kitchen windowsi

ololo11 [35]

Answer:

512 ounces

Step-by-step explanation:

there are 16 ounces in a pound so you multiply 4 x 16= 64 ounces for the 4 pounds. Since Dustin had 8 pots you multiply 64 x 8= 512 ounces.

6 0

2 years ago

A stereo store is offering a special price on a complete set ofcomponents (receiver, compact disc player, speakers, cassette dec

Korvikt [17]

Answer:

Step-by-step explanation:

(a)

The number of receivers is 5.

The number of CD players is 4.

The number of speakers is 3.

The number of cassettes is 4.

Select one receiver out of 5 receivers in $5C_1$ ways.

Select one CD player out of 4 CD players in $4C_1$ ways.

Select one speaker out of 3 speakers in $3C_1$ ways.

Select one cassette out of 4 cassettes in $4C_1$ ways.

Find the number of ways can one component of each type be selected.

By the multiplication rule, the number of possible ways can one component of each type be selected is,

The number of ways can one component of each type be selected is

$=5C_1*4C_1*3C_1*4C_1\\\\=5*4*3*4\\\\=240$

Part a

Therefore, the number of possible ways can one component of each type be selected is 240.

(b)

The number of Sony receivers is 1.

The number of Sony CD players is 1.

The number of speakers is 3.

The number of cassettes is 4.

Select one Sony receiver out of 1 Sony receivers in ways.

Select one Sony CD player out of 1 Sony CD players in ways.

Select one speaker out of 3 speakers in ways.

Select one cassette out of 4 cassettes in $4C_1$ ways.

Find the number of ways can components be selected if both the receiver and the CD player are to be Sony.

By the multiplication rule, the number of possible ways can components be selected if both the receiver and the CD player are to be Sony is,

Number of ways can one components of each type be selected

$=1C_1*1C_1*3C_1*4C_1\\\\=1*1*3*4\\\\=12$

Therefore, the number of possible ways can components be selected if both the receiver and the CD player are to be Sony is 12.

(c)

The number of receivers without Sony is 4.

The number of CD players without Sony is 3.

The number of speakers without Sony is 3.

The number of cassettes without Sony is 3.

Select one receiver out of 4 receivers in 4C_1 ways.

Select one CD player out of 3 CD players in 3C_1 ways.

Select one speaker out of 3 speakers in 3C_1 ways.

Select one cassette out of 3 cassettes in 3C_1 ways.

Find the number of ways can components be selected if none is to be Sony.

By the multiplication rule, the number of ways can components be selected if none is to be Sony is,

$=4C_1*3C_1*3C_1*3C_1\\\\=108$

[excluding sony from each of the component]

Therefore, the number of ways can components be selected if none is to be Sony is 108.

(d)

The number of ways can a selection be made if at least one Sony component is to be included is,

= Total possible selections -Total possible selections without Sony

= 240-108

= 132

Therefore, the number of ways can a selection be made if at least one Sony component is to be included is 132.

(e)

If someone flips the switches on the selection in a completely random fashion, the probability that the system selected contains at least one Sony component is,

$= \text {Total possible selections with at least one Sony} /\text {Total possible selections}$

= 132 / 240

= 0.55

The probability that the system selected contains exactly one Sony component is,

$= \text {Total possible selections with exactly one Sony} /\text {Total possible selections}$ $\frac{1C_1*3C_1*3C_1*3C_1+4C_11C_13C_13C_1+4C_13C_13C_13C_1}{240} \\\\=\frac{99}{240} \\\\=0.4125$

Therefore, if someone flips the switches on the selection in a completely random fashion, then is the probability that the system selected contains at least one Sony component is 0.55.

If someone flips the switches on the selection in a completely random fashion, then is the probability that the system selected contains exactly one Sony component is 0.4125.

6 0

2 years ago

4. k/6 = 8/16 5. 5.4/7 = 27/h 6. 8/11 = 4/y+2

kupik [55]

Answer:

4. $k=3\\$