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3 years ago

What is the place value of the number 9 in 328.095?

Mathematics

2 answers:

lukranit [14]3 years ago

7 0

Answer:The hundredths place

Step-by-step explanation:

brilliants [131]3 years ago

6 0

Answer:

The tenths place

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If −5e^x^2 y=x and y(−5)=0, find y′(−5) by implicit differentiation

xenn [34]

Differentiate:
-5×2(xe^x^2)y-5e^x^2dy/dx=1
-5e^x^2(2xy+y´)=1
When x=-5, y=0:
-5e²⁵y´=1, so y´(-5)=-eˉ²⁵/5

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3 years ago

Help please it would be very much appreciated!!!!!! <br> thank you

sammy [17]

Should be a!

26^2 - 22^2 = 192, take the sqrt of that C:

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3 years ago

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5. Solve the problem by writing an inequality. A club decides to sell T-shirts for $12 as a fund-raiser. It costs $20 plus $8 pe

kykrilka [37]

T=the amount of shirts
12t-20-8t=100
12t-8t=120
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So they must sell at least 30 shirts to make $100.

8 0

3 years ago

Find the horizontal asymptote off of x equals quantity 3 x squared plus 3x plus 6 end quantity over quantity x squared plus 1.

vaieri [72.5K]

Answer:

y = 3

Step-by-step explanation:

y = (3x² + 3x + 6) / (x² + 1)

The power of the numerator and denominator are equal, so as x approaches infinity, y approaches the ratio of the leading coefficients.

y = 3/1

3 0

4 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago