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3 years ago

Please just please help.....I’m desperate......

Mathematics

2 answers:

Gennadij [26K]3 years ago

8 0

Answer:

Step-by-step explanation:

the closest square numbers to 43 are 36=6^2 and 49=7^2 so sqrt43 is between 6 and 7

Lostsunrise [7]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

The value V of a Porsche 718 Cayman that is t years old can be modeled by V(t) = 420,000(0.965)t (a) What would be worth the car

nordsb [41]

Answer:

A. V (2) = $ 391,114.50

B. t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days

Step-by-step explanation:

Given formula:

V(t) = 420,000 * (0.965)^t

A. What would be worth the car′s worth in 2 years?

V(t) = 420,000 * (0.965)^t

We replace t by 2, as follows:

V(2) = 420,000 * (0.965)²

V(2) = 420,000 * 0.931225

V (2) = $ 391,114.50

B. In how many years will the car be worth $325,000?

V(t) = 420,000 * (0.965)^t

We replace V(t) by 325,000, as follows:

325,000 = 420,000 * (0.965)^t

325,000/420,000 = (0.965)^t

0.77381 = (0.965)^t

t = log 0.965(0.7738)

t = log 0.7738/log 0.965

t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days

0.2 years = 0.2 * 12 = 2.4 months

0.4 months = 0.4 * 30 = 12 days

7 0

3 years ago

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi

boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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".

Description in words of the parameter p

$p$ represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that have experienced bullying

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula

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

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.