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3 years ago

2.1 x 10^5 +4.3 x 10^4

Mathematics

1 answer:

Alla [95]3 years ago

4 0

Answer:

253,000

Step-by-step explanation:

$10^{5} = 100000$
$10^{4} = 10000$
Plug the above answers in: 2.1 × 100,000 + 4.3 × 10,000
2.1 × 100,000 = 210,000
Plug 210,000 in: 210,000 + 4.3 × 10,000
4.3 × 10,000 = 43,000
Plug 43,000 in: 210,000 + 43,000
210,000 + 43,000 = 253,000

I hope this helps!

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The greater of two numbers is 6 more than 4 times the smaller. Their sum is 41. FInd the numbers.

schepotkina [342]

Answer:

7 and 34

Step-by-step explanation:

let x and y be the 2 numbers with x > y

The 2 equations to solve are

x = 4y + 6 → (1)

x + y = 41 → (2)

Substitute x = 4y + 6 into (2)

4y + 6 + y = 41 , that is

5y + 6 = 41 ( subtract 6 from both sides )

5y = 35 ( divide both sides by 5 )

y = 7

Substitute y = 7 into (1)

x = 4(7) + 6 = 28 + 6 = 34

The 2 numbers are 7 and 34

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Answer:

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Step-by-step explanation:

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geniusboy [140]

Answer:

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3 years ago

Karen made 175 quarts of punch for her son and 11.3 quarts for

Alenkinab [10]

Answer:

186.3 quarts of punch

Step-by-step explanation:

175+11.3=186.3

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3 years ago

A recent survey showed that in a sample of 100 elementary school teachers, 15 were single. In a sample of 180 high school teache

trapecia [35]

Answer:

By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single

Step-by-step explanation:

let p1 be the proportion of elementary teachers who were single

let p2 be the proportion of high school teachers who were single

Then, the null and alternative hypotheses are:

$H_{0}$ : p2=p1

$H_{a}$ : p2>p1

We need to calculate the test statistic of the sample proportion for elementary teachers who were single.

It can be calculated as follows:

$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of high school teachers who were single ( $\frac{36}{180} =0.2$ )
p is the proportion of elementary teachers who were single ( $\frac{15}{100} =0.15$ )

N is the sample size (180)

Using the numbers, we get

$\frac{0.2-0.15}{\sqrt{\frac{0.15*0.85}{180} } }$ ≈ 1.88

Using z-table, corresponding P-Value is ≈0.03

Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)

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3 years ago