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3 years ago

AB = 15, BC = 10, and CD= 7. Find the length DA.

Mathematics

1 answer:

harkovskaia [24]3 years ago

4 0

Answer:

451. No, the angles are wrong.

Step-by-step explanation:

450. AB = 15, BC = 10, and CD= 7. Find the length DA.

This cannot be done without additional information about the sort of figure that ABCD is. If these are points on a line segment, we need to know their order. If these are points on a quadrilateral, we need to know its description in more detail.

If these are points ordered ABCD on a line, then AD = 15+10+7 = 32.

451. See the attached figure. BPQD is not a parallelogram: BCQ is not a straight line. (The internal angles of a pentagon are 108°, but would need to be 120° for BCQ to be a straight line, making BP parallel to DQ.) Instead, BPQD is an isosceles trapezoid.

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Write as an equation: Alice, Barbara, and Carol are sisters. Alice is 3 years younger than Barbara, and Barbara is 5 years young

Dmitrij [34]

a+b+c=68

b-3=a

c-5=b

now just solve the system of equations, substitue so that there are only b's in the equation:

a+b+c=68

(b-3) + b + (b+5) = 68

3b=66

b=22

Therefore Barbara is 22

4 0

3 years ago

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

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 students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

7 0

3 years ago

Total of y and five divided by x

Ede4ka [16]

Answer:

5y/x

Step-by-step explanation:

this is all i can write as an equation boo

4 0

3 years ago

Read 2 more answers

Please I really need an answer to this.