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Fynjy0 [20]
3 years ago
9

Start at 158. Create a pattern that adds 8. Stop when you have 5 numbers

Mathematics
2 answers:
Arisa [49]3 years ago
7 0
158,166,174,182,190,198




Alekssandra [29.7K]3 years ago
3 0
\bf \begin{array}{cll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
1&158\\
2&158+8\\
&166\\
3&166+8\\
&174\\
4&174+8\\
&182\\
5&182+8\\
&190
\end{array}
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If DA = 6 and MD = 4, what is the length of DK?
Leto [7]

Answer:

DK = 9

Step-by-step explanation:

In triangle AMD,

h² = p² + b²

or, h² = 6² + 4²

or, h²= 36 + 16

so, h² = 52

so, AM² = 52

Take x as reference angle,

cos²x = 16/52

Now,

In triangle, AMK,

Taking x as reference angle,

cos²x = b²/h²

cos²x = AM²/MK²

or, cos²x = 52/MK²

Now,

cos²x = 16/52 = 52/MK²

or,

16/52 = 52/MK²

or, 16MK² = 2704

or, MK² = 2704/16

or, MK² = 169

so, MK = 13

Now,

DK = MK - MD

or, DK = 13 - 4

so, DK = 9

3 0
3 years ago
What is the percentage of 15 out of 75
Archy [21]

 \frac{15}{75}= \frac{3}{15}= \frac{1}{5}= \frac{20}{100}



The answer is 20%


4 0
3 years ago
Read 2 more answers
A sample size 25 is picked up at random from a population which is normally
Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

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

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

6 0
3 years ago
How do you do this?<br> Because i don't know how to do it :&gt;
Sindrei [870]

Answer:

£ 7.95

Step-by-step explanation:

Let the total money = £ x

Money spend on drinks and cake = 1.45 + 1.20 = £ 2.65

Money left with her = (2/3) of x

So, money spend = (1/3) of x

\dfrac{1}{3}*x= 2.65\\\\\\x=2.65*3\\\\

x = £ 7.95

7 0
2 years ago
Which values are solutions to the inequality below?<br> Check all that apply.
Ludmilka [50]
I think the answer is -100
5 0
3 years ago
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