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1 year ago

How do I do this? 50 points, please answer

Mathematics

1 answer:

xz_007 [3.2K]1 year ago

5 0

Check the picture below.

now, the midsegment of a triangle, besides being half the base, we have to bear in mind that is a "mid" segment, is a segment that lies half-way along the segments it touches.

so for the triangle above, well, the midsegment DE is cutting the segment BC it two equal halves, BE and EC, that means that $\stackrel{x}{BE}~~ = ~~\stackrel{8}{EC}$ .

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A blue die and a red die are thrown. Each die has six sides, labeled 1 through 6. The dice are fair, so each outcome is equally

lidiya [134]

Answer:0.138

Step-by-step explanation:

Given

A blue die and a red die is rolled once such that each face is marked with different number

Favorable outcome of getting 6 is

(5,1) , (1,5) , (4,2) , (2,4) , (3,3)

Total outcomes =36

Probability of getting a sum of 6 is $=\frac{5}{36}=0.138$

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3 years ago

The angle measures of a triangle are shown in the diagram.

oksano4ka [1.4K]

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Step-by-step explanation:

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3 years ago

Fermium-253 has a half life of three days. A sample contains 400 g. How long will it take for the sample to decay to 100 g?

NemiM [27]

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3 years ago

Find the outlier of the set data 24,18,26,20,18,24,4,24

DIA [1.3K]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or

tigry1 [53]

Answer-

The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%

Solution-

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

$P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}$

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}$

$P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67$

Putting these values,

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{0.15}{0.67} =0.224=22.4\%$

5 0

3 years ago