All categories

3 years ago

The temperature is 5 degrees about 0 degrees. If it drops 8 degrees, what is the temperature after it drops?

Mathematics

2 answers:

vredina [299]3 years ago

8 0

Answer:

ÇÇÇÇÇÇÇÇÇÇÇÇÇÇÇÇ

Step-by-step explanation:

ycow [4]3 years ago

6 0

Answer:

-3

Step-by-step explanation:

You might be interested in

There are 80 sixth graders at Wilson Middle School. Only 70% of the sixth graders will attend the morning assembly. How many six

zysi [14]

1 student will attend the assembly in the morning

4 0

3 years ago

Read 2 more answers

During one week, David worked 16 hours at his normal hourly rate and

asambeis [7]

Answer:

43.49

Step-by-step explanation:

let his normal hourly rate be x

16x+7x/2=848

39x/2=848

x=848 ÷ 39/2

=43.48717

6 0

3 years ago

Read 2 more answers

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

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 for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

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

The population proportion have the following distribution

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

Solution to the problem

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 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.