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3 years ago

How far up a wall will an 11-meter ladder reach, if the foot of the ladder is 4 meters away from the base of the wall?

Mathematics

1 answer:

Serhud [2]3 years ago

7 0

Answer: approximately 10.4 feet

Step-by-step explanation:

You might be interested in

What's the answer? and how to do this?

castortr0y [4]

A. 79
The inscribed angle is always 1/2 of the arc.
In this case angle RST is the inscribed angle and arc RT is the arc
The arc is 158

Inscribed angle=.5arc
Inscribed angle=.5(158)
Inscribed angle=79

6 0

3 years ago

Read 2 more answers

laren had 3 liters of water. she poured the water into 7 glasses equally. how much water was there in each glass.

LenKa [72]

Answer:

My answer is 14.5 ounces in each cup.

Step-by-step explanation:

4 0

3 years ago

Anyone, please help me please answer my question because I have to pass this tomorrow morning:(

Wittaler [7]

Step-by-step explanation:

If you need help with how I got my answer, you can ask me.

$\frac{2yx {}^{ - 4} }{(x {}^{ - 4}y {}^{4}) {}^{3} \times 2x {}^{ - 1}y {}^{ - 3} }$

$\frac{2yx {}^{ - 4} }{x {}^{ - 12}y {}^{12} \times 2x {}^{ - 1}y {}^{ - 3} }$

$\frac{2y x {}^{ - 4} }{2x {}^{ - 13} y {}^{ - 9} }$

$= x {}^{9} y {}^{ - 8}$

$= \frac{x {}^{9} }{y {}^{8} }$

12.

$( \frac{2u {}^{ 4} }{ - u {}^{2}v {}^{ - 1} \times 2uv {}^{ - 4} } ) {}^{ - 1}$

$( \frac{2u {}^{4} \times 1 }{ - 2 {u}^{3}v {}^{ - 5} } ) {}^{ - 1}$

$( - u v {}^{ 5} ) {}^{ - 1}$

$= - u {}^{ - 1} v {}^{ - 5} = - \frac{ 1}{uv {}^{5} }$

13.

$- \frac{2m {}^{4}n {}^{ - 1} }{( - m {}^{2}n {}^{ - 2}) {}^{ - 1} \times - nm {}^{ - 3} }$

$- \frac{2m {}^{4}n {}^{ - 1} }{ - m {}^{ - 2} n {}^{2} \times - nm {}^{ - 3} }$

$- \frac{2m {}^{4} n {}^{ - 1} }{m {}^{ - 5}n {}^{3} } = - 2m {}^{9} n {}^{ - 4}$

$= - \frac{2m {}^{9} }{n {}^{4} }$

14.

$( \frac{x {}^{3}y {}^{ - 4} }{ - {x}^{2} y {}^{ - 3} \times yx {}^{0} } ) {}^{3}$

$( \frac{x {}^{3}y {}^{ - 4} }{ - {x}^{ - 2}y {}^{ - 2} } ) {}^{3}$

$( - {x}^{5} y {}^{ - 2} ) {}^{3} = - x {}^{15} y {}^{ - 6}$

$- \frac{x {}^{15} }{ {y}^{6} }$

15.

$- \frac{yx {}^{4} \times - {y}^{3} z { }^{ - 4} }{(z {y}^{2} ) {}^{4} }$

$- \frac{ - y {}^{4} x {}^{4} z {}^{ - 4} }{ {z}^{4} y {}^{8} } = - y {}^{4} x {}^{4} z {}^{ - 8} = - \frac{(xy) {}^{4} }{ {z}^{8} }$

16.

$\frac{h {}^{7}j {k}^{4} }{4h {}^{4} } = \frac{1}{4} h {}^{3} = \frac{h {}^{3} jk {}^{4} }{4}$

6 0

3 years ago

In a random sample of 80 people, 52 consider themselves as baseball fans. Compute a 92% confidence interval for the true proport

Thepotemich [5.8K]

Answer:

k0020

Step-by-step explanation:

4 0

3 years ago

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

4 years ago