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3 years ago

Pls help me for question no.4

Mathematics

1 answer:

Natalija [7]3 years ago

4 0

Answer:

Area of the composite figure = 75.25 cm²

Step-by-step explanation:

Question (4). Given figure is a composite figure having,

(1). Right triangle STU

(2). A kite PSUV

(3). A trapezoid PQRS

Now we will calculate the area of each figure.

(1). Area of the right triangle = $\frac{1}{2}(\text{ST})(\text{TU})$

= $\frac{1}{2}(3.5)(3)$

= 5.25 cm²

(2). Area of the kite PSUV = $\frac{1}{2}(\text{Diagonal 1})(\text{Diagonal 2})$

= $\frac{1}{2}(\text{PU})(\text{SV})$

= $\frac{1}{2}(\text{TS+RQ})(\text{SV})$

= $\frac{1}{2}(3.5+7)(6)$ [Since SV = 2 × 3 = 6 cm]

= $3\times 10.5$

= 31.5 cm²

(3). Area of the trapezium = $\frac{1}{2}(b_1+b_2)(h)$ [Where $b_1$ and $b_2$ are the bases and h is the distance between the bases]

= $\frac{1}{2}[(7-3)+7](7)$

= $\frac{77}{2}$

= 38.5 cm²

Total area of the given figure = 5.25 + 31.5 + 38.5

= 75.25 cm²

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In a simple random sample of 14001400 young people, 9090% had earned a high school diploma. Complete parts a through d below.

ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has increased.

Step-by-step explanation:

Let p = proportion of young people who had earned a high school diploma.

A sample of n = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

$\hat p=0.90$

(a)

The standard error for the estimate of the percentage of all young people who earned a high school diploma is given by:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Compute the standard error value as follows:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=\sqrt{\frac{0.90(1-0.90)}{1400}}\\$

$=0.008$

Thus, the standard error for the estimate of the percentage of all young people who earned a high school diploma is 0.0080.

(b)

The margin of error for (1 - α)% confidence interval for population proportion is:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

$=1.96\times 0.0080\\=0.01568\\\approx1.6\%$

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

$CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)$

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has increased, the hypothesis is defined as:

H₀: The percentage of young people who earn high school diplomas has not increased, i.e. p = 0.80.

Hₐ: The percentage of young people who earn high school diplomas has not increased, i.e. p > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of p, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has increased.

8 0

3 years ago

Please help....There are 42 boys and girls participating in an essay-writing competition. Of the competitors, 21 are in seventh

almond37 [142]

The probability of an 8th grader winning is : 14/42 which reduces to 1/3

the simulation would be : # 3 <====

because u will have 6 equal sections on ur spinner...with each section representing 7 people
one section representing 9th graders...so their is 7 of them
2 sections representing 8th graders....so their is 7 * 2 = 14 of them
and 3 sections representing 7th graders....7 * 3 = 21 of them

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victus00 [196]

Answer: quotient is 2x^2 + 10x - 5

Solution:

The first polynomial is miswritten.

The right one is: 2x^3 + 4x^2 - 35x + 15.

So, the division is [2x^3 + 4x^2 - 35x + 15] / (x - 3)

The synthetic division uses the coeffcients and obviate the letters, but you have to be sure to respect the place of the coefficient.

So, in this case it is:

3 | 2      4    -35    15
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    2     10     - 5      0

So, the quotient is 2x^2 + 10x - 5, and the remainder is 0.

I like to show it in this other way:

   | 2     4     -35     15
   |
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3 |       +6    +30     -15
--------------------------------
    2     10    - 5         0

Of course they are the same coefficients and the answer continue being quotien 2x^2 + 10x - 5, remainder 0.

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What is the solution of the equation (x -- 5)2 + 3(x - 5) + 9 = 0? Use u substitution and the quadratic formula to solve

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Answer:

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Answer:

Step-by-step explanation:

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3 years ago

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