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3 years ago

AC = 13 in the image shown. Is BC tangent to circle A?

Mathematics

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

Yes, because 52 + 122 = 132.

Step-by-step explanation:

A line through a point on a circle is tangent at the point if it is perpendicular to the radius drawn to the point of tangency. The Pythagorean Theorem only works for right triangles.

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A landscaper is selecting two trees to plant. He has five to choose from. Three of the five are deciduous and two are evergreen.

ICE Princess25 [194]

Answer:

The probability that he chooses trees of two different types is 30%.

Step-by-step explanation:

Given that a landscaper is selecting two trees to plant, and he has five to choose from, of which three of the five are deciduous and two are evergreen, to determine what is the probability that he chooses trees of two different types must be performed the following calculation:

3/5 x 2/4 = 0.3

2/5 x 3/4 = 0.3

Therefore, the probability that he chooses trees of two different types is 30%.

4 0

2 years ago

Read 2 more answers

9.21 <img src="https://tex.z-dn.net/?f=9.21%20%5Cdiv%200.4%20%3D%20" id="TexFormula1" title="9.21 \div 0.4 = " alt="9.21 \di

Whitepunk [10]

Let us do it by converting the 0.4 to a fraction, it has one decimal, so we'll use one zero at the denominator.

and 9.21 to a fraction as well, two decimals, thus two zeros at the denominator then,

$\bf 9.21\div 0.4\\\\
-------------------------------\\\\
0.\underline{4}\implies \cfrac{04}{1\underline{0}}\implies \cfrac{4}{10}\implies \cfrac{2}{5}\qquad \qquad \qquad \qquad 9.\underline{21}\implies \cfrac{921}{1\underline{00}}\\\\
-------------------------------\\\\
9.21\div 0.4\implies \cfrac{921}{100}\div \cfrac{2}{5}\implies \cfrac{921}{100}\cdot \cfrac{5}{2}\implies \cfrac{921}{20}\cdot \cfrac{1}{2}\implies \cfrac{921}{40}
\\\\\\
23\frac{1}{40}\implies 23.025$

3 0

3 years ago

Help please!!! will mark brainiest later

Aloiza [94]

1. 1/2

2. The Y axis

3. 1 right and 1 down (1, -1)

7 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

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

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

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

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

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

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

$p_v =P(t_{68}>4.31)=0.000022$