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Elina [12.6K]
1 year ago
7

Identifying a Valid Sample Natasha wants to find out if the neighborhood supports lowering the speed limit on the street in fron

t of her school. Which is a valid sample for the survey? students in Natasha’s social studies class parents who live in a town 15 miles away from Natasha’s school the first 5 drivers who drive past Natasha’s school one morning thirty residentsIdentifying a Valid Sample Natasha wants to find out if the neighborhood supports lowering the speed limit on the street in front of her school. Which is a valid sample for the survey?
A. students in Natasha’s social studies class parents who live in a town 15 miles
B.away from Natasha’s school the first 5 drivers who drive past Natasha’s C.school one morning thirty residents who live within a 2-miles radius of D.Natasha’s school who live within a 2-miles radius of Natasha’s school
Mathematics
2 answers:
nekit [7.7K]1 year ago
7 0

D - Thirty Residents Who Live Within a 2-Miles Radius Of Natasha’s School

blsea [12.9K]1 year ago
3 0

Answer:

thirty residents who live within a 2-miles radius of Natasha’s school

Step-by-step explanation:

You might be interested in
32.15 is written correctly in EXPANDED FORM using FRACTIONS
marshall27 [118]

Answer:

Step-by-step explanation:

: Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).

5 0
1 year ago
To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of μ=40 , and the treatment is
Westkost [7]

Answer:

Step-by-step explanation:

a)

Test statistic:

t=\frac{35-40}{\sqrt{\frac{32}{8} } }

t=-2.5

df=8-1=7

critical, t ,values=+/-2.365

here test statistic lie in rejection region,that why null hypothesis fails  

so Yes, its significant.

b)

Test statistic:

t=\frac{35-40}{\sqrt{\frac{72}{8} } }

t=-1.67

df=8-1=7

critical, t ,values=+/-2.365

c)

sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.

5 0
1 year ago
A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

A' = 0

This gives:

0 = -512y^{-2} + 2

Collect Like Terms

512y^{-2} = 2

Multiply through by y^2

y^2 * 512y^{-2} = 2 * y^2

512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

Length = 10

Width = 16 + 4

Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0
1 year ago
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
1 year ago
Harper knows he is 50 yards from school. The map on his phone shows that the school is 34 inch from his current location. How fa
laila [671]

To solve this problem you must apply the proccedure shown below:

1. We know that 1 yard is 36 inches, therefore, 50 yards expressed in inches is:

(50)(36)=1800in

2. If he is 50 yards from school and the map shows that the school is 34 inches from his current location, when it shows 3 inches the real distance is:

\frac{(1800in)(3in)}{34in}=158.82in

3. If you can to express it in yards:

\frac{158.82}{36} =4.41 yd

Therefore, the answer is: 158.82 inches or 4.41 yards.

8 0
2 years ago
Read 2 more answers
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