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2 years ago

(04.02 MC)

Mathematics

1 answer:

diamong [38]2 years ago

5 0

Answer: D, Put it in stocks

Step-by-step explanation: It’s not B, C, and A because Donald said that he doesn’t trust banks or markets. And the A wouldn’t be the answer either cause safes are from banks

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Sonny has $75 to spend. The purchase he wants to make requires$93. If he borrows the extra money that he needs, how much does he

lidiya [134]

All you gotta do is subtract both of the numbers to get how much you need.

93 - 75 = 18

We can double check to see if it's right by using addition

18 + 75 = 93

Sonny needs to borrow 18$ to meet his required amount :D

7 0

3 years ago

5/8 divided by 8 3/10

almond37 [142]

Answer:

25/332

Step-by-step explanation:

4 0

2 years ago

For a school picnic, leslie ordered a box of fresh-baked gingerbread cookies and sugar cookies. She got 60 cookies in all. 15 of

OlgaM077 [116]

Answer:

Pretty sure 25%.

Step-by-step explanation:

First, you would need to see how many times 60 goes into 100, because a percent is ALWAYS over 100. You should get an answer of 1.66666667. Then you if you multiply it to one of the numbers then you have to do it to the other number. So, you would do 15 times 1.66666667. You would get 25. So, that would be 25 over 100 or 25/100. And 25 over 100 is 25%.

I hope I helped! :) Sorry if it's wrong :(

5 0

2 years ago

How to solve logarithmic equations as such

Serga [27]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5$

$\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}$

5 0

3 years ago

Compute the conditional probabilities from the two-way frequency table.

Natali5045456 [20]

Answer:

The table is show clearly in the figure attached.

P(boy if favourite activity is swimming) = 8/17 = 0.47

P(girl if favourite activity is sport) = 7/27 = 0.26

P(girl if favourite activity is reading) = 4/6 = 0.67

P(boy if favourite activity is sport) = 20/27 = 0.74

P(favourite activity is swimming if a girl) = 9/20 = 0.45

P(favourite activity is reading if a boy) = 2/30 = 0.07

P(favourite activity is swimming if a boy) = 8/30 = 0.27

P(favourite activity is reading if a girl) = 4/20 = 0.2

6 0

2 years ago