Which of the following mathematical expressions is equivalent to the verbal expression "a number, X,

An object is thrown from a height of 5 in. After 2 s, the object reaches a maximum height of 9 in, and then it lands back on the

Marta_Voda [28]

Answer:

y=-(x-2)^2+9

Step-by-step explanation:

Hope this helps

7 0

2 years ago

Mr.Walden wrote the expression. He asked his students to write an equivalent expression of simplified form.

mafiozo [28]

Answer:

option D

Brianna

Step-by-step explanation:

Given in the question the expression wrote by Mr.Walden

$\frac{p^{-5} }{q^{0} }$

To write the simplified form of this expression we will use negative rule of exponent

$b^{-n}$ = 1 / $b^{n}$

$q^{-5} = \frac{1}{q^{5} }$

so,

$\frac{1}{q^{5} x q^{0} }$

$\frac{1}{q^{5} p^{0} }$

Only Brianna wrote right simplification of the expression written by Mr.Walden

3 0

2 years ago

Jessie has a deck of 52 regular playing cards and a bag of six marbles. In the bag, there are two blue marbles, three green marb

denpristay [2]

Answer:

Im so sorry i dont know

Step-by-step explanation:

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7 0

2 years ago

Read 2 more answers

A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot

Damm [24]

Answer:

a) $p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) $\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.

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".

Part a

Description in words of the parameter p

$p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

$\hat p$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

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

The population proportion have the following distribution

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

Part b

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.

$\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

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 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.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545$

$0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591$

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.

5 0

2 years ago

A 100-gallon barrel, initially half-full of oil, develops a leak at the bottom. Let A(t) be the amount of oil in the barrel at t

gayaneshka [121]

Answer:

the mathematical model is : $-\frac{1}{A}= kt - \frac{1}{50}$

Step-by-step explanation:

Given that:

Let A(t) to be the amount of oil in the barrel at time t.

However; Suppose that the amount of oil is decreasing at a rate proportional to the product of the time elapsed and the amount of oil present in the barrel.

Then,

$\frac{dA}{dt} \ \alpha \ A^2$