Round 6435 to the nearest thousands

Which expression is equivalent to 3 + 5(2x + 9) - 3

Sladkaya [172]

Answer:

12

Step-by-step explanation:

5 0

3 years ago

Please help quickly.Thanks :)

Klio2033 [76]

Multiply 2x-5y= -21 by 3 to make it 6x-15y= -63
Multiply 3x-3y= -18 by -5 to make it -15x+15y=90

This cancels the y’s out which leaves us with

6x=-63
&
-15x=90

x for 6x=-63 equals - 10.5 so x is - 10.5 and for -15x=90, x= -6

Then you plug in x into any equation you’d like to find y.

Let’s plug in - 10.5 into 6x... equation.

6(- 10.5)-15y=-63

63-15y= -63
-63 -63

-15y=0
y=0 and x= - 10.5. When you plug in this values it makes the equation true!

But the correct answer is the first one north. Sorry if I’m doing too much hahah

If I’m confusing here’s the right answer...

6x-15y= -63
-15x+15y=90

5 0

3 years ago

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Interne

Dvinal [7]

Answer:

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

Step-by-step explanation:

Given that in a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week. The mean of the sample observations was 12.9 hours.

$H_0: \bar x =12.7\\H_a: \bar x >12,7$

(Right tailed test at 5% level)

Mean difference = 0.2

Std error = $\frac{6}{\sqrt{1000} } \\=0.1897$

Z statistic = 1.0540

p value = 0.145941

since p >alpha we do not reject H0.

b. Do not reject H0. We do not have convincing evidence that the mean weekly time spent using the Internet by Canadians is greater than 12.7 hours.

5 0

3 years ago

If a*b=a^2-b^3, then what is the value of 3*-2

mezya [45]

I think it’s -6 but I’m not sure

3 0

2 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$