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2 years ago

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Mr. Rijo bought 49 bags of soil one week and 28 bags of soil the next week. Which expression correctly applies the distributive

property to show equivalent expressions for the number of bags of soil he bought? 49 + 28 = 4 (7 + 7) 49 + 28 = (7) (7) + (14) (2) 49 + 28 = (7) (14) + (7) (7) 49 + 28 = 7 (7 + 4) plz i need help

1 answer:

Montano1993 [528]2 years ago

6 0

Answer:

I don't know the answer bro but please follow me bro

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In a random sample of 64 people, 48 are classified as 'successful.' Determine the sample proportion of 'successful' people.

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The sample proportion of 'successful' people is $\frac{3}{4}$ .

Step-by-step explanation:

The sample consist of 64 people and 48 of them are 'successful'. Hence, the proportion of 'successful' people is:

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$N$ - People that forms the sample, dimensionless.

$n$ - People classified as 'successful', dimensionless.

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3 years ago

Read 2 more answers

Which point on the graph of g(x)=(1/5)^x? HELPP

Cloud [144]

Answer:

(-1,5) and $(3, \frac{1}{125})$ are points on the graph

Step-by-step explanation:

Given

$g(x) = \frac{1}{5}^x$

Required

Determine which point in on the graph

To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))

A. (-1,5)

x = -1

Substitute -1 for x in $g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-1}$

Convert to index form

$g(x) = 1/(\frac{1}{5})$

Change / to *

$g(x) = 1*(\frac{5}{1})$

$g(x) = 5$

This satisfies (-1,5)

<em>Hence, (-1,5) is on the graph</em>

<em></em>

B. (1,0)

x = 1

Substitute 1 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^1$

$g(x) = \frac{1}{5}$

<em>(1,0) is not on the graph because g(x) is not equal to 0</em>

C. $(3, \frac{1}{125})$

x = 3

Substitute 3 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^3$

Apply law of indices

$g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}$

$g(x) = \frac{1}{125}$

This satisfies $(3, \frac{1}{125})$

<em>Hence, </em> $(3, \frac{1}{125})$ <em> is on the graph</em>

<em></em>

D. $(-2, \frac{1}{25})$

x = -2

Substitute -2 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-2}$

Convert to index form

$g(x) = 1/(\frac{1}{5}^2)$

$g(x) = 1/(\frac{1}{5}*\frac{1}{5})$

$g(x) = 1/(\frac{1}{25})$

Change / to *

$g(x) = 1*(\frac{25}{1})$

$g(x) = 25$

This does not satisfy $(-2, \frac{1}{25})$

<em>Hence, </em> $(-2, \frac{1}{25})$ <em> is not on the graph</em>

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