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2 years ago

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Tomorrow, Mrs. Wendel's class will be using toothpicks for a science project. Each student must use at least 5 toothpicks for th

e project.
Mrs. Wendel knows that there is already a bag of 50 toothpicks in the class storage room. She plans to buy x additional toothpicks this afternoon to make sure her class will have enough.

If her class has 29 students, which number sentence represents this situation?

A.

1/29x+1 21/29 ≥ 5

B.

1/29+25/29 ≥ 5

C.

1/29+25/29 ≤ 5

D.

1/29x+1 21/29 ≤ 5

1 answer:

Amiraneli [1.4K]2 years ago

7 0

Answer:

I'm pretty sure A but let me know if I'm wrong

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Can I have help on this question, please?

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Nikki drew a rectangle with a perimeter of 18 units on a coordinate grid. Two of the vertices were (4, –3) and (–1, –3). What co

Talja [164]

Answer:

There are two possible solutions for the other two vertices of the rectangle:

(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)

Step-by-step explanation:

Geometrically speaking, the perimeter of a rectangle ( $p$ ) is:

$p = 2\cdot b + 2\cdot h$ (1)

Where:

$b$ - Base of the rectangle.

$h$ - Height of the rectangle.

Let suppose that the base of the rectangle is the line segment between (4, -3) and (-1, -3). The length of the base is calculated by Pythagorean Theorem:

$b = \sqrt{[(-1)-4]^{2}+[(-3)-(-3)]^{2}}$

$b = 5$

If we know that $p = 18$ and $b = 5$ , then the height of the rectangle is:

$2\cdot h = p-2\cdot b$

$h = \frac{p-2\cdot b}{2}$

$h = \frac{p}{2}-b$

$h = 4$

There are two possible solutions for the other two vertices of the rectangle:

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2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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2 years ago

Alaskan Salmon are fished extensively to serve in restaurants. However, there are limits to how many and the size of fish which

TEA [102]

Answer:

The salmon size is less than equal to 10.28 inches represent bottom 25% of all salmon.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12.5 inches

Standard Deviation, σ = 3.3

We are given that the distribution of salmon size is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.25

$P( X < x) = P( z < \displaystyle\frac{x - 12.5}{3.3})=0.25$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 12.5}{3.3} = -0.674\\\\x = 10.2758\approx 10.28$

Thus, the salmon size is equal to or less than 10.28 inches, they are considered small and young and represent bottom 25% of all salmon.

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Answer:

87.5% percent chance.

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