How Many Different Angle Measures Can Be Found Among The Eight Angles Formed When Two Parallel Lines Are Cut By A Transversal?

If an architect wants to draw a plan for a house that is 75 ft by 80 ft, and his paper is 10 in. by 17 in., what is the best sca

drek231 [11]

The best scale factor to make use of every inch of the paper would be that 1 inch represents 7.5 feet.

7 0

2 years ago

The figure shows a square, a regular pentagon and a regular hexagon put together. What is the measure of ∠BAC

Elan Coil [88]

Answer:

42

Step-by-step explanation:

One angle of a regular pentagon = 108°

One angle of a regular hexagon = 120°.

One angle of a square = 90°

∴ ∠BAC = 360 – (108 + 20 + 90) 360 – 318 = 42°

8 0

1 year ago

Find the area of the<br> parallegram

sweet [91]

Answer:

To find the area, multiply the base by the height. The formula is: A = B * H where B is the base, H is the height, and * means multiply. The base and height of a parallelogram must be perpendicular.

Step-by-step explanation:

hope it helps

please mark brainliest

3 0

2 years ago

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.

Kobotan [32]

Answer:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=3.7$ represent the mean for athletes

$\bar X_{NA}=4.1$ represent the mean for non athletes

$s_{A}=1.1$ represent the sample standard deviation for athletes

$s_{NA}=1.3$ represent the sample standard deviation for non athletes

$n_{A}=32$ sample size for the group 2

$n_{NA}=45$ sample size for the group 2

$\alpha=0.01$ 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 population mean for athletes is lower than the population mean for non athletes, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{NA}\geq 0$

Alternative hypothesis: $\mu_{A} - \mu_{NA}< 0$

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:

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

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:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

P value

We need to find first the degrees of freedom given by:

$df=n_A +n_{NA}-2=32+45-2=75$

Since is a one left tailed test the p value would be:

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

5 0

2 years ago

When determining how well an observed set of frequencies fits an expected set of frequencies, what is the test statistic

sergij07 [2.7K]

Answer: $\chi^2$ test statistic.

Step-by-step explanation:

The chi-square ( $\chi^2$ ) test is a non-parametric statistical test(also known as Goodness of fit test) which is used to determine if a distribution of observed frequencies differs from the expected frequencies.

Chi-square statistic uses either nominal (categorical) or ordinal level data.

Hence, When determining how well an observed set of frequencies fits an expected set of frequencies, the test statistic is $\chi^2$ test statistic.

6 0

2 years ago