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3 years ago

Can someone tell me what 24.5 x 3.2 is

Mathematics

2 answers:

Scrat [10]3 years ago

6 0

Answer:

the answer for you question is 78.4

omeli [17]3 years ago

3 0

Answer: 78.4

Step-by-step explanation:

above^

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Potato cause 0.32 per pound how much does 0.75 pounds of sweet potatoes cost

Semmy [17]

Answer: .24

You get this by multiplying .75 by .32 or by dividing .32 by four and multiplying by 3

5 0

3 years ago

What is the difference between 2a + 5 and 2(a + 5)? Give an English expression to describe each of

Sergio [31]

Answer:

without brackets, the two is only multiplying the a. with brackets, you distribute the 2, making the equation 2a+10

Step-by-step explanation:

6 0

3 years ago

What is the simplified answer to this? thank you :)<br><br><img src="https://tex.z-dn.net/?f=9%20%5Csqrt%7B245%7D%20" id="TexFor

Svetradugi [14.3K]

Answer:

$63 \sqrt[ \: ]{5}$

Step-by-step explanation:

$9 \sqrt{245}$

$9 \sqrt{ {7}^{2} \times 5 }$

$9 \times 7 \sqrt{5}$

$63 \sqrt{5}$

6 0

2 years ago

Number 3?? I can’t figure it out

monitta

divide the total cost by the number of hinges:

Mark's hardware: $28 / 8 = $3.50 per hinge

Steve's Supply: $39 / 11 = $3.45 per hinge.

$3.45 is less than $3.50, so the answer would be C.

7 0

2 years ago

Read 2 more answers

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel

laiz [17]

Answer:

0.03 is the probability that for the sample mean IQ score is greater than 103.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

$= 1 - P(z \leq 1.875)$

Calculation the value from standard normal z table, we have,

$P(x > 103) = 1 - 0.970 =0.03= 3\%$

0.03 is the probability that for the sample mean IQ score is greater than 103.

7 0

2 years ago