How would you write 3.5 x 10^5 in standard form ?

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!

shepuryov [24]

Answer:

4

Step-by-step explanation:

5 0

3 years ago

The label on a 1-gallon jug of fruit juice states that the serving size is 8 oz. How many servings are in the jug?

Elodia [21]

Answer: 16 fl. oz.

Step-by-step explanation:

There are 128 fl. oz. in one gallon. Divide that by the serving size, 8 fl. oz., and you get 16.

4 0

3 years ago

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2455 subjects randomly select

Genrish500 [490]

Answer: 0.355,0.405)

Step-by-step explanation:

Given : Significance level : $\alpha=1-0.99=0.01$

Number of subjects (n) = 2455

Number of surveys returned = 931

The probability of surveys get return will be :-

$p=\dfrac{931}{2455}=0.379226069246\approx0.38$

The confidence interval for proportion is given by :-

$p\pm z_{\alpha/2}\times\sqrt{\dfrac{p(1-p)}{n}}$

$0.38\pm z_{0.005}\times\sqrt{\dfrac{0.38(1-0.38)}{2455}}\\\\=0.38\pm(2.576)0.0098\\\approx0.38\pm0.025=(0.355,0.405)$

Hence, the 99% confidence interval for the proportion of returned surveys : (0.355,0.405)

5 0

3 years ago

. Mr. Lorenzo gave his 2 sons $50 to buy a cooler. The total cost for the cooler is $44. Mr. Lorenzo told his sons that they cou

Butoxors [25]

Answer:

Step-by-step explanation:

Given that the cost of cooler is $44

money given to the two sons $50

let S represent the amount the two son will have to share equally

the change is 50-44=6

=$6

Then s=6/2=$3

applying with the expression directly we have

S = (50 - 44)/2

S = (6)/2

S=$3

7 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

3 years ago

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