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ad-work [718]
3 years ago
9

Equation- a(10x+6)=b+40x. What is the value of A and B

Mathematics
1 answer:
Jobisdone [24]3 years ago
5 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

a(10x + 6) = 40x + b

10ax + 6a = 40x + b

Thus :

10ax = 40x

x s simplifies

10a = 40

Divide sides by 10

\frac{10a}{10}  =  \frac{40}{10}  \\

a = 4

_________________________________

And also must :

6a = b

b = 6a

b = 6(4)

b = 24

_________________________________

So :

a = 4

b = 24

Done...

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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Divide a line of 1.8m into three parts in the ratio of 5:7:8​
kozerog [31]

Answer:

0.45 m, 0.63 m, 0.72 m

Step-by-step explanation:

Let the three parts of the line be 5x, 7x and 8x

Therefore,

5x + 7x + 8x = 1.8

20x = 1.8

x = 1.8/20

x = 0.09

5x = 5*0.09 = 0.45 m

7x = 7*0.09 = 0.63 m

8x = 8*0.09 = 0.72 m

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3 years ago
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3 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

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".

The population proportion have the following distribution

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

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
3 years ago
Brainlist thingy and 20
Flauer [41]

Answer:

Nolan takes 100s

Step-by-step explanation:

x = x - 15x/100

x = 0.85x

x = 85/0.85

x = 100s

7 0
3 years ago
Read 2 more answers
Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of
forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

n = 120, \pi = \frac{58}{120} = 0.4833

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

8 0
2 years ago
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