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3 years ago

The parallelogram shown below has an area of 140 units^2. The base is twelve and the side is 10 help this is so hard.

Mathematics

1 answer:

faltersainse [42]3 years ago

5 0

Answer:

height = 11.67 units

Step-by-step explanation:

A parallelogram is a quadrilateral with 2 pairs of parallel sides. This means opposites sides are parallel to each other. Opposite angles and sides are congruent.

From your question the base of the parallelogram is given as 12 units and the sides is 10 units. The area of the parallelogram is given as 140 units². What is left is the height of the parallelogram.

Using the formula for finding the area of a parallelogram, the height can be found.

area of a parallelogram = base × height

base = 12 units

height = ?

area = 140 unit²

area of a parallelogram = base × height

let

h = height

140 = 12h

divide both sides by 12

h = 140/12

h = 11.6666666667 units

height = 11.67 units

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A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha

ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% 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 students received a passing grade = 77%

n = sample of tests = 40

p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u>99% confidence interval for p</u> = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ , $0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ ]

= [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

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Answer:

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