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4 years ago

Karen is trying to decide which home phone company to use. Phone Home was a monthly charge of

Mathematics

1 answer:

lys-0071 [83]4 years ago

3 0

Answer:

Is this a joke?

Step-by-step explanation:

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A fish and game warden captures and tags 66 deer in a local forest. A month later he captures 1oo deer and finds that 17 of them

NemiM [27]

Should be A. 149
100-17=83
83+66=149

4 0

4 years ago

Question Find the sum. 1/3(9−6m)+1/4(12m−8)

Umnica [9.8K]

The sum of this is =1+m

3 0

3 years ago

What is 312/17 in simplest form.

My name is Ann [436]

312/17
= (306+6)/ 17
= 306/17+ 6/17
= 18+ 6/17
= 18 6/17

The final answer is 18 6/17~

7 0

3 years ago

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?

Phantasy [73]

Answer:

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Step-by-step explanation:

Given : $\cos (180^{\circ}-q)=-\cos q$

We have to write which identity we will use to prove the given statement.

Consider $\cos (180^{\circ}-q)=-\cos q$

Take left hand side of given expression $\cos (180^{\circ}-q)$

We know

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Comparing , we get, a= 180° and b = q

Substitute , we get,

$\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}$

Also, we know $\sin 180^{\circ}=0$ and $\cos 180^{\circ}=-1$

Substitute, we get,

$\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0$

Simplify , we get,

$\cos (180^{\circ}-q)=-\cos (q)$

Hence, use difference identity to prove the given result.

7 0

3 years ago

Read 2 more answers

You are studying the relationship between the history of early childhood trauma (categorized as trauma/no trauma) and weight in

jeka57 [31]

Answer:

You can decide to reject the Null Hypothesis at p < 0.05.

Step-by-step explanation:

In this case, the relationship between the history of early childhood trauma and weight in adulthood are being studied.

The categories of childhood trauma are:

Trauma and No trauma.

So, number of rows is, r = 2.

The categories of weight in adulthood are:

Low to normal, Overweight, and Obese

So, number of columns is, c = 3.

The distribution of the statistic $\chi^{2}$ is chi-square with (r - 1)(c - 1) degrees of freedom, where r represents the number of rows in the two-way table and c represents the number of columns.

Compute the degrees of freedom as follows:

$df = (r-1)(c-1)=(2-1) \times (3-1) = 2$

It is provided that the calculated value of Chi-square test statistic is 18.26.

Compute the p-value of the test as follows:

$p-value=P(\chi^{2}_{2}\geq 18.26)\\=0.000108\\\approx0.00011$

Decision rule:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

The significance level is, α = 0.05.

p-value = 0.00011 < α = 0.05.

The null hypothesis will be rejected at 5% level of significance.

Thus, the complete statement is:

"You can decide to reject the Null Hypothesis at p < 0.05."

4 0

3 years ago