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Dvinal [7]
2 years ago
15

What is the simplified form of StartFractio negative 24 m Superscript 5 Baseline n Superscript 4 Baseline Over 8 m Superscript n

egative 7 Baseline n Superscript negative 2 Baseline End Fraction? Assume m not-equals 0, n not-equals 0. Start-fraction 3 Over m Superscript 35 Baseline n Superscript 8 Baseline End-fraction Negative Start-fraction 3 n squared Over m squared End-fraction Negative 3 m Superscript 12 Baseline n Superscript 6 Baseline 3 m Superscript 35 Baseline n Superscript 8 Baseline
Mathematics
2 answers:
NNADVOKAT [17]2 years ago
5 0

Answer:

Option C on E2020

Step-by-step explanation:

just finished the test

beks73 [17]2 years ago
3 0

Answer:

-3m^12n^6

Step-by-step explanation:

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A quality control officer randomly checking weights of pumpkin seed bags being filled by an automatic filling machine. Each bag
GarryVolchara [31]
From 0g to 494.7g are rejected as well as those heavier than 505.3grams.

You calculate this by subtracting and adding 5.3 to 500
7 0
2 years ago
Read 2 more answers
A social media platform states that a social media post from a marketing agency has 7 hashtags, on average. A digital marketing
dimaraw [331]

Answer:

a. The average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags.

Step-by-step explanation:

A social media platform states that a social media post from a marketing agency has 7 hashtags, on average.

This means that at the null hypothesis, we test if the mean is 7, that is:

H_0: \mu = 7

A digital marketing specialist studying social media advertising believes the average number of hashtags used in a post from a marketing agency is different than the number stated by the social media platform.

Keyword is different, so at the null hypothesis, we test if the mean is different of 7, that is:

H_1: \mu \neq 7

Thus, the correct answer is given by option a.

4 0
2 years ago
A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot
Damm [24]

Answer:

a) p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) \hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

Description in words of the parameter p

p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

\hat p represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

z_{\alpha/2} represent the critical value for the margin of error

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545

0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

5 0
2 years ago
What is 10.2719 rounded to the nearest hundreth?
Paladinen [302]
Find the number in the hundredth place
7
7
and look one place to the right for the rounding digit
1
1
. Round up if this number is greater than or equal to
5
5
and round down if it is less than
5
5
.
10.27
4 0
2 years ago
Read 2 more answers
A freight company has shipping orders for two products. The first product has a unit volume of 10 cu ft, and it weighs 50 lbs. T
Lady_Fox [76]

Answer:

116 units of the first product

380 units of the second product

Step-by-step explanation:

Product 1 has a unit volume of  10 cu ft

Product 2 has a unit volume of 3 cu ft

The truck has 2300 cu ft of space

Product 1 weighs 50 lbs

Product 2 weighs 40 lbs

The truck can carry 21000 lbs

Let X be the units of product 1

Let Y be the units of product 2

The given information can be expressed as:

10X+3Y=2300...(1)

50X+40Y=21000...(2)

Solving the system of equations:

10X+3Y=2300...(1)

10X=2300-3Y

X=(2300-3Y)/10

Substituting X in (2) we have:

50X+40Y=21000

50[(2300-3Y)/10]+40Y=21000

50[(2300/10)-(3Y/10)]+40Y=21000

50[230-(3Y/10)+40Y=21000

11500-(150Y/10)+40Y=21000

11500-15Y+40Y=21000

11500+25Y=21000

25Y=21000-11500

25Y=9500

Y=380

Substituting Y in (1) we have:

10X+3Y=2300...(1)

10X+3(380)=2300

10X+1140=2300

10X=2300-1140

10X=1160

X=116

So 116 units of the first product and 380 units of the second product can be transported in a single shipment with one truck.

8 0
2 years ago
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