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2 years ago

Please help with this I am so confused

Mathematics

1 answer:

musickatia [10]2 years ago

8 0

7) A
8) B

These are the answers to the question when I put it into a calculator

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Can someone help me with this?

jekas [21]

Answer:

$\huge\purple{ x^2 +(y+2)^2 = 4}$

Step-by-step explanation:

Circle is centred at (0, -2)

-> h = 0 & k = - 2

radius of the circle (r) = y-coordinate of the center = - 2

Equation of circle in standard form is given as:

$(x-h)^2 +(y-k)^2 = r^2$

Plugging the values of h, k and r in the above equation, we find:

$(x-0)^2 +[y-(-2)]^2 = (-2)^2$

$\implies\huge\orange{ x^2 +(y+2)^2 = 4}$

This is the required equation of circle.

6 0

1 year ago

Write 6(9+2k)= -3 in written form

Alchen [17]

Answer:

27k+3=0

Step-by-step explanation:

Write in standard form.

this way?

4 0

2 years ago

8. A paper bag contains one pink crayon, four yellow crayons, three black crayons and two red crayons. What is the probability o

vivado [14]

Answer:

C 1/2

Add all the crayons together and then see how many red and black crayons there are. which in this case it's half. That is the probability

8 0

2 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago

Select all the ways that describe the graph of the function in interval 2.(multiple choice)

Sergio039 [100]

B. Slope is negative
and
C. Function decreasing

6 0

2 years ago