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3 years ago

Please help with 1. I am confused

Mathematics

2 answers:

DaniilM [7]3 years ago

6 0

Add 8 1/2 every box because every half if 8 1/2 so the first box would be 17 because 8 1/2 + 8 1/2 = 17

d1i1m1o1n [39]3 years ago

4 0

It would be 9, 9 1/2, 10, and 10 1/2. I think.

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Multiply. (3.5x10^-5) (3x10^-10) express your answer in scientific notation.

Len [333]

Answer:

5.69034 × 1011

Step-by-step explanation:

Decimal

1234000000

Scientific Notation

1.234 x 10^9

×10^

1st Number

1.234

×10^

Operation

2nd Number

5.678

×10^

3 0

3 years ago

11. Set up the rational work application problem. Shelby and Mary started a beading

Arada [10]

The first one might be b or c

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3 years ago

14.4 A Book and a Cake A book costs $45. Sales tax on the book is 7%. Write two different expressions that represent the final c

snow_tiger [21]

Answer: 1. $45*1.07 | $45(1.07) 2. (0.85)*32 represent the final price of the cake in dollars because (0.85) is the 15% off in decimals.

Step-by-step explanation:

1 if you multiply 45 by 1.07 it will be 48.15 and that's the final cost of the book, same with $45(1.07) is just another way to represent the multiplication.

2 As I said the 0.85 represent the 15% off, and when you multiply it will be less.

3 0

3 years ago

Grades on a standardized test are known to have a mean of 1000 for students in the US. The test is administered to 453 randomly

Mademuasel [1]

Answer:

a: 1003 < µ < 1023

b: See attached photo for answers

c: see below

d and e: See second attached photo

f: See below

Step-by-step explanation:

a. to find a confidence interval, first find the error with the formula,

E = Z(σ/√n) where Z is the value of the z-score that relates to the confidence level, σ is the standard deviation of the population (or of the sample if the population standard deviation isn't given), and n is the sample size

For this problem, E = 1.96(108/√453) = 9.95 = 10

Now add and subtract that value form the sample average...

x - E < µ < x + E becomes 1013 - 10 < µ 1013 + 10

simplifies to 1003 < µ < 1023

b. See attached photo for hypothesis test results...

c. H0: µ = 1013

Ha: µ > 1013 (claim)

They want to test if it's greater than their first attempt, so this claim becomes the alternate hypothesis. We're testing against their first attempt, that's why µ = 1013, not 1000

d and e. See second attached photo for hypothesis test results...

f. There is enough evidence to support the claim that the students second attempt was an improvement from their first attempt. They happened to take a 2 hour tutoring course between tests. This doesn't prove that the tutoring was effective thought. It very well might have been effective, but also, the students taking the test again means they've already seen the test and know what to expect, which could have also boosted their scores.

8 0

4 years ago

A random sample of 21 observations is used to estimate the population mean. The sample mean and the sample standard deviation ar

stepladder [879]

Answer:

a. CI=[128.79,146.41]

b. CI=[122.81,152.39]

c. As the confidence level increases, the interval becomes wider.

Step-by-step explanation:

a. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula;

$\bar X\pm \ z\frac{s}{\sqrt{n}},$

where:

$s$ is the sample standard deviation
$z$ is the s value of the desired confidence interval

we then calculate our confidence interval as:

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]$

Hence, the 95% confidence interval is between 128.79 and 146.41

b. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula in a above;

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]$

Hence, the variable's 99% confidence interval is between 122.81 and 152.39

c. -Increasing the confidence has an increasing effect on the margin of error.

-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.

-The 99% Confidence interval is the most appropriate to use in such a case.

7 0

4 years ago