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3 years ago

Solve each equation.1) 10 = b + 4 - 76

Mathematics

1 answer:

Anna71 [15]3 years ago

7 0

Answer:the answer is 82 photo math helps a lot with these type of questions

Step-by-step explanation:

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Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=900 and the number of succ

oee [108]

Answer:

An 99% confidence interval of the given proportion

(0.355 , 0.385)

Step-by-step explanation:

Given sample size n= 900

the number of successes in the sample is x=333

The proportion P = $\frac{x}{n} = \frac{333}{900} = 0.37$

Q = 1-P =1 - 0.37 = 0.63

Confidence interval:-

99% of confidence interval zα = 2.93

$(P - z_{\alpha } \sqrt{\frac{PQ}{n} } , P + z_{\alpha } \sqrt{\frac{PQ}{n} })$

$(0.37 - 2.93 \sqrt{\frac{0.37(0.63}{900} } ,0.37 +2.93 \sqrt{\frac{0.37(0.63}{900} } })$

(0.37 - 0.015 , 0.37 + 0.015)

(0.355 , 0.385)

Conclusion:-

An 99% Confidence interval (0.355 , 0.385)

7 0

3 years ago

pls help mee I dont k ow this and if anyone is smort enough can you help me with the 9ther questions (5 i total my last redo)

melisa1 [442]

Answer:

Step-by-step explanation:

I think that is the answer

7 0

3 years ago

Read 2 more answers

Let v represent the value of a bank acount, in thousands of dollars, t yearsaftertye account is opend at the bank

Liula [17]

Answer:

could you explain it more

Step-by-step explanation:

4 0

2 years ago

I need help with this

Assoli18 [71]

On a given line, (on one side) there are a total of 180°

if one line in Problem #3 is bisected by a line, with one half X and the other 120°,

do 180° (the total) minus 120° which=60°

now the hard part, that line that bisected the first line is bisecting a line that is parallel to your second line, the one with <5 and <6

this means that the big angle formed in the first one with 120° is the same angle as in the second line, leaving <5 as 120°
which means <6 is 60°, like in the top part of the problem. You're basically flipping the top line upside down, I hope it helps.

7 0

3 years ago

Occasionally an airline will lose a bag. Suppose a small airline has found it can reasonably model the number of bags lost each

julia-pushkina [17]

Answer:

a) The probability that the airline will lose no bags next monday is 0.1108

b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227

c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2

Step-by-step explanation:

The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

$P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}$

$P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108$

The probability that the airline will lose no bags next monday is 0.1108.

b) Note that $P(X \in \{0,1,2\} = P(X=0) + P(X=1) + P(X=2)$ . And

$P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227$

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.

c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.

6 0

3 years ago

Solve each equation.1) 10 = b + 4 - 76​

Solve each equation.1) 10 = b + 4 - 76