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3 years ago

6.7 more than the product of 5 and n

2 answers:

6 0

The product of 5 and n is 5n. Since the expression is 6.7 more than this product, we add 6.7 to 5n. the expression becomes:

5n+6.7

bagirrra123 [75]3 years ago

4 0

6.7+5n

more means to add

and product means to multiply

so its 6.7+5n

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Simplify this expression <br> (1/3 + 11/12)×2/3

alisha [4.7K]

1/3 = 4/12

(4/12 + 11/12) * 2/3

= 15/12 * 2/3

15/ 12 can also be written as 5/4

5/4 * 2/3 = 10/12

10/12 can be simplified into 5/6

Therefore, the answer is 5/6

5 0

3 years ago

F(n) =7n -11 <br> f(1)<br> f(2)

UkoKoshka [18]

Answer:

- 4 and 3

Step-by-step explanation:

Given

f(n) = 7n - 11

To evaluate f(1) and f(2) substitute n = 1 and n = 2 into f(n), that is

f(1) = 7(1) - 11 = 7 - 11 = - 4

f(2) = 7(2) - 11 = 14 - 11 = 3

5 0

3 years ago

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lara31 [8.8K]

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

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

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: