(3 3/8 – 1 1/5) + 1 7/8

I don’t understand this at all

Mnenie [13.5K]

Answer:

B = 61.0

Step-by-step explanation:

Since this is a right angle, we can use trig functions

sin theta = opp side/ hypotenuse

sin B = 7/8

Take the inverse of each side

sin ^-1 ( sin B) = sin ^ -1 (7/8)

B =61.04497563

To the nearest tenth

B = 61.0

5 0

3 years ago

How many ways are there to distribute the cookies if there are six choco- late cookies, six oatmeal cookies and six red velvet c

tangare [24]

Answer:

6

Step-by-step explanation:

7 0

2 years ago

Please answer correctly ! Will mark brainliest answer !!!

Andreyy89

Answer:

The answer is the first one.

6 0

2 years ago

Determine the solution to the equation x²6x=-13

Citrus2011 [14]

Answer:

x = $\frac{\sqrt[3]{468}}{6}$

or x =~ -1.29399

Step-by-step explanation:

Calculate the product. x^2 * 6x = 6x^3

divide both sides by 6. x^3 = -13/6

Take the root of both sides. x = $\frac{\sqrt[3]{468}}{6}$

3 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago