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3 years ago

6th grade math , help me please :)

Mathematics

2 answers:

Charra [1.4K]3 years ago

7 0

Answer:

(a) 3:5

(b) 15

Step-by-step explanation:

Well first we need to create a ratio for the 5 shades squares and 3 non+shaded squares.

It is asking for unshaded first so our ratio will be,

3:5

(a) 3:5

So if there is 9 unshaded how many shaded is there.

Well we can just make the following ratio 9:x

So 9/3 = 3

Meaning 3*5 = 15

So x = 15

(b) 15

ArbitrLikvidat [17]3 years ago

7 0

Answer:

a) 3/5

b) 15

Step-by-step explanation:

(a)

unshaded: 3

shaded: 5

unshaded to shaded: 3/5

(b)

There are now 3 unshaded squares. You need 9 unshaded squares, so you need to have three times as many total squares as you have now.

Add two more lines just like the given line.

Each new line will have 5 shaded and 3 unshaded squares.

Now you have a total of 9 unshaded and 15 shaded squares.

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Will give Brainliest! if right!

Bogdan [553]

There are 11 letters in total.
T & B
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3 years ago

Help and explain tyty:)

Kamila [148]

I hope this one help you

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3 years ago

Use the equation s=5q+8 to find the value of s when q=1.

Norma-Jean [14]

Answer:

s=13

Step-by-step explanation:

s=5q+8

plug in numbers

s=5(1)+8

solve 5*1

s=5+8

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7 0

2 years ago

What is the factored form of this expression 4x^2 -12x+9

Anon25 [30]

Answer:

(2x - 3) ^2

Step-by-step explanation:

Factor using the perfect square rule.

7 0

3 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

8 0

2 years ago