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2 years ago

Can someone help me on B and C

Mathematics

1 answer:

Elis [28]2 years ago

6 0

B. If you look at the graph and go over 5cm and up 3m, you can see that that point is not near the others, so it can't be accurate according to the data on your graph. C. If you go over 3.2 cm on your graph, go up to where the other points are, and you will see that the most likely height would be about 4.7m. Try to imagine a line going through your graph around where the points are, that might help you see where other points would be. I hope that makes sense let me know if you need anything else I'm always happy to help :)

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Simplify the expression. (5+i)(2-7i)

scZoUnD [109]

Answer:

17-33i

Step-by-step explanation:

Multiply using the foil method than combined the real and imaginary parts of the expression

5 0

2 years ago

A retailer buys a coat in bulk at a unit price of $45. The coat is marked up to 28% to sell in the store. What is the price of t

inessss [21]

The answer is $57.60

(The original price) 45+ 12.60 (28% of 45) = 57.60

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2 years ago

(1/27)^x-6=27 Could I please have a step by step as well? Much appreciated!

Andrei [34K]

Answer:

$x = 7$

Step-by-step explanation:

1) Use Division Distributive Property: (x/y)^a = x^a/y^a.

$\frac{1}{ {27}^{x - 8} } = 27$

2) Multiply both sides by 27^x - 8.

$1 = {27} \times 27^{1 + x - 8}$

3) Use the product rule: x^a x^b = x^a+b.

$1 = {27}^{1 + x - 8}$

4) Simplify 1 + x - 8 to x - 7.

$1 = {27}^{x - 7}$

5) Use Definition of Common Logarithm: b^a = x if and only if logb (x) = a.

$log_{27}1 = x - 7$

6) Use Change of Base Rule.

$\frac{ log_1 }{ log{27} } = x - 7$

7) Use rule of 1: log 1 = 0.

$\frac{0}{ log_{27}} = x - 7$

8) Simplify 0/log_27 to 0.

$0 = x - 7$

9) Add 7 to both sides.

$7 = x$

10) Switch sides.

$x = 7$

Therefor, the answer is x = 7.

5 0

2 years ago

If x degree and 115 degree are co exterior angles find the vaule of x degree

Lesechka [4]

Answer:

65 degrees

Step-by-step explanation:

An exterior angle form a linear pair with an interior angle.

A linear pair of angles must add up to 180 degrees

180-115 = 65

3 0

2 years ago

A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that

mafiozo [28]

Answer:

(a) The significance level of the test is 0.002.

(b) The power of the test is 0.3487.

Step-by-step explanation:

We are given that a coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that the probability is not 0.5.

The test rejects the null hypothesis if either 0 or 10 heads are observed.

Let p = probability of obtaining head.

So, Null Hypothesis, $H_0$ : p = 0.5

Alternate Hypothesis, $H_A$ : p $\neq$ 0.5

(a) The significance level of the test which is represented by $\alpha$ is the probability of Type I error.

Type I error states the probability of rejecting the null hypothesis given the fact that the null hypothesis is true.

Here, the probability of rejecting the null hypothesis means we obtain the probability of observing either 0 or 10 heads, that is;

P(Type I error) = $\alpha$

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

Also, the event of obtaining heads when a coin is thrown 10 times can be considered as a binomial experiment.

So, X ~ Binom(n = 10, p = 0.5)

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

$\binom{10}{0}\times 0.5^{0} \times (1-0.5)^{10-0} +\binom{10}{10}\times 0.5^{10} \times (1-0.5)^{10-10}$ = $\alpha$

$(1\times 1\times 0.5^{10}) +(1 \times 0.5^{10} \times 0.5^{0})$ = $\alpha$

$\alpha$ = 0.0019

So, the significance level of the test is 0.002.

(b) It is stated that the probability of heads is 0.1, and we have to find the power of the test.

Here the Type II error is used which states the probability of accepting the null hypothesis given the fact that the null hypothesis is false.

Also, the power of the test is represented by (1 - $\beta$ ).

So, here, X ~ Binom(n = 10, p = 0.1)

$1-\beta$ = P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true)

$1-\beta$ = $\binom{10}{0}\times 0.1^{0} \times (1-0.1)^{10-0} +\binom{10}{10}\times 0.1^{10} \times (1-0.1)^{10-10}$

$1-\beta$ = $(1\times 1\times 0.9^{10}) +(1 \times 0.1^{10} \times 0.9^{0})$

$1-\beta$ = 0.3487

Hence, the power of the test is 0.3487.

3 0

2 years ago