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3 years ago

Write a math problem that includes multiplying decimals.Explain how you know where to place the decimal in the product.

Mathematics

1 answer:

disa [49]3 years ago

6 0

Answer:

0.1 x 2 = 0.2

Step-by-step explanation:

when you multiply, you move the decimal once because there is one decimal place.

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Tammy baked a batch of 20 cupcakes. Daniella baked

ElenaW [278]

Multiplication is the process of multiplying. The total number of cupcakes that are produced by the 3 girls is 44.

<h3>What is multiplication?</h3>

Multiplication is the process of multiplying, therefore, adding a number to itself for the number of times stated. For example, 3 × 4 means 3 is added to itself 4 times, and vice versa for the other number.

Given that Tammy baked a batch of 20 cupcakes and Daniella baked 4/5 as many cupcakes as Tammy. Therefore, the number of cupcakes made by Daniella is,

Number of cupcakes made by Daniella

= (4/5)×(Number of cupcakes made by Tammy)

= (4/5) × 20

= 16

Now, if Miko baked 1/2 as many cupcakes as Daniella. Therefore, the number of cupcakes made by Miko is,

Number of cupcakes made by Miko

= (1/2) × (Number of cupcakes made by Daniella)

= (1/2) × 16

= 8

Further, the total number of cupcakes that are produced are,

Number of cupcakes = 20 + 16 + 8 = 44

Hence, the total number of cupcakes that are produced by the 3 girls is 44.

Learn more about Multiplication here:

brainly.com/question/14059007

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8 0

1 year ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago

9. To make lemonade you need 30 ounces of water and

love history [14]

Answer:

Step-by-step explanation:

Hopefully this helps

6 0

3 years ago

~**Will mark brainliest **~<br> For the correct answers to all three questions

aleksandr82 [10.1K]

$\dfrac{42}{65}\cdot\dfrac{25}{36}\cdot\dfrac{26}{49}=\dfrac{2\cdot3\cdot7}{5\cdot13}\cdot\dfrac{5\cdot5}{2\cdot2\cdot3\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\dfrac{7}{13}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\\\\\\=\dfrac{1}{1}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{1}{7}=\dfrac{5}{21}$

$\dfrac{21}{32}\cdot\dfrac{39}{120}\cdot\dfrac{40}{65}=\dfrac{21}{32}\cdot\dfrac{3\cdot13}{2\cdot2\cdot2\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot2\cdot5}{5\cdot13}=\\\\\\=\dfrac{3\cdot7}{32}\cdot\dfrac{13}{2\cdot2\cdot2\cdot5}\cdot\dfrac{2\cdot2\cdot2}{13}=\dfrac{21}{32}\cdot\dfrac{1}{5}\cdot\dfrac{1}{1}=\dfrac{21}{160}$

$\dfrac{15}{90}\cdot\dfrac{36}{75}\cdot\dfrac{27}{42}=\dfrac{3\cdot5}{2\cdot3\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot3\cdot3}{3\cdot5\cdot5}\cdot\dfrac{3\cdot3\cdot3}{2\cdot3\cdot7}=\\\\\\=\dfrac{1}{2\cdot3}\cdot\dfrac{2\cdot2\cdot3}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{1}\cdot\dfrac{2}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{25}\cdot\dfrac{9}{7}=\dfrac{9}{175}$

6 0

3 years ago

Find the probability of rolling a twelve first and then a three when a pair of dice is rolled twice?

yawa3891 [41]

Total number of possible outcomes = 36

Total number of outcomes with a sum of 12 = 1
{6,6)

Total number of outcomes with a sum of 3 = 2
{1,2}{2,1}

P(12 first, then 3) = (1/36)(2/36) = 1/648

Answer: The probability is 1/648

6 0

3 years ago

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