All categories

2 years ago

6. Kathleen needs 240 valid signatures on a petition to get a crosswalk installed on her street. She knows

Mathematics

1 answer:

fredd [130]2 years ago

3 0

Kathleen must get 300 signatures to make sure she will have enough valid signatures for the crosswalk

<em><u>Solution:</u></em>

Given that Kathleen needs 240 valid signatures on a petition to get a crosswalk installed on her street

She knows that about 20% of the signatures will not be valid for a variety of reasons

Let "x" be the number of signatures

invalid signature = 20 % of x

valid signatures = 240

Then, according to question,

total number of signatures - invalid signatures = 240

x - 20 % of x = 240

$x - 20 \% \times x = 240\\\\x - \frac{20}{100} \times x = 240\\\\x - 0.2x = 240\\\\0.8x = 240\\\\x = 300$

Therefore, she must get 300 signatures so that she will have enough valid signatures for the crosswalk

You might be interested in

What is the maximum vertical distance between the line y = x + 56 and the parabola y = x2 for −7 ≤ x ≤ 8?

atroni [7]

-7 and 8 are the solutions to the given equation system.
Therefore, the maximum distance between the y values of the two equations must lie exactly between their points of intersection. That is on x value:
x = (-7 + 8)/2 = 0.5
The maximum distance is:
y = 0.5 + 56 = 56.5
y = 0.5² = 0.25
56.5 - 0.25 = 56.25 units

5 0

2 years ago

Round to the nearest ten thousand 905154

kolbaska11 [484]

You want to round 905,154 to the nearest ten-thousands place. The ten-thousands place in your number is shown by the bold underlined digit here:

9<em><u>0</u></em>5,154

    To round 905,154 to the nearest ten-thousands place...

    The digit in the ten-thousands place in your number is the 0. To begin the rounding, look at the digit one place to the right of the 0, or the 5, which is in the thousands place.

    Since the 5 is greater than or equal to 5, we'll round our number up by

        Adding 1 to the 0 in the ten-thousands place, making it a 1.

    and by changing all digits to the right of this new 1 into zeros.

The result is: 910,000.

So, 905,154 rounded to the ten-thousands place is 910,000.

5 0

2 years ago

PLEASE HELP ASAP, CORRECT ANSWER GETS BRAINLIEST

Fynjy0 [20]

Car value at the time of purchase $15, 250 (2012)
Rate of depreciation 7.5%
1) y = 15250 ( 1- 0.075)^x
y = 15250 (0.925)^x (x tis the number of years that passed since purchase)

2) y = 15250 * ( 0.925)^8 = 8173.42

4 0

2 years ago

Read 2 more answers

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago

How many possible values for y are there where y = cos^-1 0?

Arada [10]

Possible values for y are<u> infinite</u>

<h3>Further explanation </h3>

Trigonometry is the science of mathematics that studies the relationship between sides and angles in triangles

Given special angles of trigonometric functions, for example

$\displaystyle sin~0=0\\\\cos~30=\frac{1}{2}\sqrt{3}\\\\tan~45=1\\\\sec~45=\sqrt{2}\\\\etc.$