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3 years ago

Pleaseeee helppp answer correctly !!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

⦣ACD = 90º

Step-by-step explanation:

⦣ACD + ⦣BCD = 180º

⦣BCD = 90º

⦣ACD + 90º = 180º subtract 90º from both sides

⦣ACD = 90º

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A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determi

grandymaker [24]

The mass of the first shipment at time t is
$m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}$

The mass of the second shipment at time t is
$m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }$

At time t, the ratio of m₁ to m₂ is
$\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123$

Therefore as a percentage,
$\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%$

Answer: B. 81.2%

4 0

3 years ago

What is the value of z in the equation 2(4z − 3 − 1) = 166 − 46?

ZanzabumX [31]

In this equation z=16

7 0

3 years ago

Someone please help me with this question

zubka84 [21]

You can use pythagorean theorem and square 1.5 and 6 and those added together is the hypotenuse (the slope) squared. the answer is approximately 6.2

8 0

3 years ago

Assume the general population gets an average of 7 hours of sleep per night. You randomly select 45 college students and survey

tekilochka [14]

Answer:

a) ii. This is a left-tailed test.

b) -1.59

c) -1.301

d) i. reject null hypothesis

e) Option i) The data supports the claim that college students get less sleep than the general population.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 7 hours

Sample mean, $\bar{x}$ = 6.87 hours

Sample size, n = 45

Alpha, α = 0.10

Sample standard deviation, s = 0.55 hours

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}$

a) We use one-tailed(left) t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59$

c) Now,

$t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301$

Since,

$t_{stat} < t_{critical}$

d) We fail to accept the null hypothesis and reject it.

We accept the alternate hypothesis and conclude that mean number of hours of sleep for all college students is less than 7 hours.

e) Option i) The data supports the claim that college students get less sleep than the general population.

8 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago