Is this linear or exponential? or neither

Help! Which estimate is closest to the actual value of ( 2.99548 ) ⋅ ( 1.8342 )?

nordsb [41]

Answer:

Step-by-step explanation:

2.99548 * 1.8342

the first number rounds to 3.....the second number rounds to 2.....3 * 2 = 6...but it will be a little short of 6 because of the 1.8.....so I am going with 5.5

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jada ran 7.1 miles.....noah ran 5.123 miles

so jada ran (7.1 - 5.123) = 1.977 miles farther then noah

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greater then 3...

1.67 + 1.4 and 4.5 - 1.47

8 0

3 years ago

I NEED HELP PLEASEEE I WILL MARK YOU BRAINLIEST

Gemiola [76]

Yess what do you need help with

5 0

2 years ago

Mateo made 36 cookies with nuts. He also made 12 fewer cookies without nuts.

igor_vitrenko [27]

48 cookies in total

6 0

2 years ago

Read 2 more answers

Bob and ling are playing a number pattern game. Bob wrote the following sequence

seraphim [82]

The unknown term is 22.6

28.9
26.8
24.7
22.6
20.5

As you can see, the numbers after the decimal point descends by 1 point. 9 , 8 , 7 , 6 , 5
While the numbers before the decimal point descends by 2 points, 28 , 26 , 24 , 22 , 20

6 0

3 years ago

Read 2 more answers

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

3 years ago