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erastovalidia [21]
3 years ago
12

In the diagram, AC=BD. Show that AB=CD

Mathematics
1 answer:
Debora [2.8K]3 years ago
8 0
I believe you just have to draw 2 little lines in between AB and CD.
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There are 20 sweets in a box.
natima [27]

Answer:

probability \: of \: getting \: red \: sweets  =  \frac{x}{20}  \\ probability \: of \: getting \: yellow \: sweets \:  =  \frac{(20 - x)}{20}

5 0
3 years ago
what are the explicit equation and domain for a geometric sequence with a first term of 5 and a second term of -10
OleMash [197]
a_1=5;\ a_2=-10;\ r=a_2:a_2\to r=-10:5=-2\\\\a_n=a_1r^{n-1}\\\\\boxed{a_n=5\cdot(-2)^{n-1}}
6 0
3 years ago
Read 2 more answers
PLS HELP <br><br>*50 points*
snow_tiger [21]

Remember that there is 100 centimeters in a meter, and 1000 meters in a kilometer. If you multiply both values together (so you have a conversion of centimeters to kilometers) you get 100,000. Therefore, 100,000 centimeters makes a kilometer. Divide 320,000 by 100,000 and you get 3.2. Therefore, a centimeter would represent 3.2 kilometers.

8 0
2 years ago
Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.
algol [13]

Answer:

f\left( {2.1} \right) \approx 5.22360.

Step-by-step explanation:

The linear approximation is given by the equation

                            {f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}

Linear approximation is a good way to approximate values of f(x) as long as you stay close to the point x= a, but the farther you get from x=a, the worse your approximation.

We know that,

a=2\\f(2) = 5\\f'(x) = \sqrt{3x-1}

Next, we need to plug in the known values and calculate the value of f(2.1):

{L\left( x \right) = f\left( 2 \right) + f^\prime\left( 2 \right)\left( {x - 2} \right) }=5+\sqrt{3(2)-1}(x-2) =5+\sqrt{5}(x-2)

Then

f\left( {2.1} \right) \approx 5+\sqrt{5}(2.1-2)\approx5.22360.

6 0
3 years ago
Time spent using​ e-mail per session is normally​ distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that
liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 11, \sigma = 3

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 25, s = \frac{3}{\sqrt{25}} = 0.6

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{11.2 - 11}{0.6}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293.

X = 10.8

Z = \frac{X - \mu}{s}

Z = \frac{10.8 - 11}{0.6}

Z = -0.33

Z = -0.33 has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

Z = \frac{X - \mu}{s}

Z = \frac{11 - 11}{0.6}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 10.5

Z = \frac{X - \mu}{s}

Z = \frac{10.5 - 11}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 100, s = \frac{3}{\sqrt{100}} = 0.3

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{11.2 - 11}{0.3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

X = 10.8

Z = \frac{X - \mu}{s}

Z = \frac{10.8 - 11}{0.3}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0
3 years ago
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