All categories

3 years ago

ANSWERS TRIANGLE BFE TRIANGLE BFH TRIANGLE BFD TRIANGLE BFG

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

The answer is BFG because those angles form a triangle while the other options form pyramids

borishaifa [10]3 years ago

5 0

That would be triangle BFG answer

You might be interested in

Please help me Simplify 3✔️2 - ✔️2

MAXImum [283]

$\bf \stackrel{like~terms}{3\sqrt{2}-\sqrt{2}}\implies 2\sqrt{2}$

3 0

3 years ago

A spinner is divided into equal sections that are labeled with integer values.

Julli [10]

Answer:

1/2

Step-by-step explanation:

look at all the numbers then look at only the negatives which are 2 then count the positive ones which are 4 from those numbers you get 2/4 simplify it, and you get 1/2.

8 0

3 years ago

Activity

hodyreva [135]

Answer:

The answers to each part are:

Part A.

The quantity of apples is one-third of the quantity of grapes.

Part B.

The quantity of apples is a quarter of the quantity of strawberries.

Part C.

The number of cherries is two-elevenths of the total fruit.

Step-by-step explanation:

To identify the answer in each case, you must remember that all the parts are equal, then:

Part A.

The parts of apples are 1 and the parts grapes are 3, so if you divide the first quantity with the second quantity you obtain:

1 / 3 = 1/3 (one-third)

So, the quantity of apples is one-third of the quantity of grapes or the quantity of apples is three times smaller than the quantity of grapes.

Part B.

The parts of apples are 1 and the parts of strawberries are 4, then you must divide the first quantity with the second quantity:

1 / 4 = 1/4 (a quarter)

In this case, the quantity of apples is a quarter of the quantity of strawberries or the quantity of apples is four times smaller than the quantity of strawberries.

Part C.

First, you must add all the part of fruit:

1 part apple
1 part orange
4 parts strawberry
2 parts cherry
3 parts grape

The total of fruits is 11 parts, taking into account the quantity of cherries is 2, now you can divide the number of cherries with the total parts of fruit:

2 / 11 = 2/11 (two-elevenths)

Now, you can see the number of cherries is two-elevenths of the total fruit.

3 0

3 years ago

Please help me :)))!!!!!!

astraxan [27]

Answer:

16 × X + 3=

Step-by-step explanation:

u will get your answer

4 0

3 years ago

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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 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 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.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago