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elena-s
9 days ago
9

Sharon is assessing the popularity of two fashion magazines for her blog. The number of subscribers to magazine A and the number

of subscribers to magazine B for the years 2000 to 2004 are given in the tables below.
The number of subscribers to magazine A in 2000 is 5,000, 20,000, 25,000, or 10,000 more, or less than the number of subscribers to magazine B in 2000.

Which of the following describes the number of subscribers represented by the tables? Only the number of subscribers to magazine B is increasing. , The numbers of subscribers to both magazines are increasing. , The number of subscribers to both magazines are decreasing. , or Only the number of subscribers to magazine B is decreasing.

As the number of years since 2000 increases, the number of subscribers to only magazine B approaches infinity, only magazine A approaches zero, only magazine A approaches infinity, only magazine B approaches zero, both magazines approach negative infinity, or both magazines approach zero .

Mathematics
2 answers:
zzz [11.8K]9 days ago
6 0
1) In 2000, magazine A had 5,000 subscribers more than magazine B. 2) The available options indicate that the subscription numbers for both magazines are dropping. 3) As the years progress from 2000, the subscriber counts for both magazines approach zero.
babunello [11.3K]9 days ago
3 0
What’s the answer?? The previous response wasn’t even fully complete.
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An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th
tester [11933]

Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X is the sample mean

\mu indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In this instance, the 9% confidence interval corresponds to:

8.8104 \leq \mu \leq 11.1248

We can determine the mean using the following:

\bar X = \frac{8.8104 +11.1248}{2}= 9.9676

Additionally, the margin of error can be calculated as:

ME= \frac{11.1248- 8.8104}{2}= 1.1572

The margin of error for this situation is expressed as:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

Next, we find the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

SE = \frac{1.1572}{1.96}= 0.5904

For the 98% confidence interval, the significance corresponds to \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 with a critical value of 2.326, yielding a confidence interval of:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

8 0
29 days ago
A boy throws a baseball across a field. The ball reaches its maximum height, 18 meters, after 1 second. After approximately 2.3
PIT_PIT [11939]

Since the ball achieves a peak height of 18 meters at 1 second, it will begin descending thereafter. Therefore, we anticipate that the height will drop below 18 m after 1.5 sec.

f(x) = –10x2 + 20x + 8
f(x) = –10(1.5^2) + 20(1.5) + 8
f(x) = 15.5 m

<span> </span>

5 0
10 days ago
Read 2 more answers
A bag holds 12 red marbles, 11 green marbles, 17 blue marbles, and 5 yellow marbles. What is the probability that you will not c
Zina [12015]

¡Hola! Bienvenido a!

Vamos a sumar cuántas canicas tenemos en total.

12+11+17+5=45

Queremos hallar la probabilidad de elegir una canica que no sea azul. Observemos cuántas canicas no son azules.

12+11+5=28

Tendremos esta probabilidad sobre 48.

28/48

Al simplificar, obtenemos 7/12 o alrededor de 58.33%.

¡Espero que esto ayude!

7 0
1 month ago
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"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight
AnnZ [11946]

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

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6 0
1 month ago
Myrtle has a credit card that uses the average daily balance method. For the first 21 days of one of her billing cycles, her bal
AnnZ [11946]

Answer:

A

I encountered this question very recently.

5 0
1 month ago
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