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3 years ago

6390 In scientific notation ( no decimal)

Mathematics

1 answer:

pshichka [43]3 years ago

3 0

6.39 x 10^3 is the answer to the question

or rounded to 6 x 10^3

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1. Fill in the blanks for the rules for working with exponents:

natima [27]

1.a)add
B) substract
C) multiply
D) one

2) 1002=1.002×10^3

3)2.8×10^-3 < 2.5×10^-2 < 1.2×103 <4×103

5 0

3 years ago

Read 2 more answers

Lou has an account with $10,000 which pays 6% interest compounded annually. If to that account, Lou deposits $5,000 at the begin

katovenus [111]

Answer:

Option d. $22154 is the right answer.

Step-by-step explanation:

To solve this question we will use the formula $A=P(1+\frac{r}{n})^{nt}$

In this formula A = amount after time t

P = principal amount

r = rate of interest

n = number of times interest gets compounded in a year

t = time

Now Lou has principal amount on the starting of first year = 10000+5000 = $15000

So for one year $A=15000(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 15000(1+.06)^{1}$

$= 15000(1.06)$ = $15900

After one year Lou added $5000 in this amount and we have to calculate the final amount he got

Now principal amount becomes $15900 + $ 5000 = $20900

Then putting the values again in the formula

$A=20900(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 20900(1+.06)^{1}$

$= 20900(1.06)=22154$

So the final amount will be $22154.

3 0

3 years ago

What is 96/126 in simplest form

Elina [12.6K]

$\frac{96}{126}\\\\96\ and\ 126\ are\ even\ numbers\ divisible\ by\ 2\\\\\frac{96:2}{126:2}=\frac{48}{63}\\\\48\ and\ 63\ are\ divisible\ by\ 3,\ because\ 4+8=12(divisible\ by\ 3)\\and\ 6+3=9(divisible\ by\ 3)\\\\\frac{48:3}{63:3}=\boxed{\frac{16}{21}}\leftarrow answer$

5 0

3 years ago

Read 2 more answers

Una caja tiene 60 bombones eba se comio 1/5 bombones y ana 1/2 bombones cuantos bombones se comio eba y ana

DedPeter [7]

Responder:

42 bombones

Explicación paso a paso:

Dado que :

Número de bombones = 60

Cantidad consumida por Eba = 1/5 * 60 = 12

Cantidad comida por Ana = 1/2 * 60 = 30

Cantidad de chocolate que comieron Eba y Ana:

(12 + 30) = 42

De ahí que Ana y Eba se comieran 42 bombones

7 0

3 years ago

In 2008, data from the Center for Disease Control revealed that 28.5% of all male teenagers, aged 18-19 and attending U.S. colle

m_a_m_a [10]

Answer:

a) H0 : u = 28.5%

H1 : u < 28.5%

b) critical value = - 1.645

c) test statistic Z= - 1.41

d) Fail to reject H0

e) There is not enough evidence to support the professor's claim.

Step-by-step explanation:

Given:

P = 28.5% ≈ 0.285

X = 210

n = 800

$p' = \frac{X}{n} = \frac{210}{800} = 0.2625$

Level of significance = 0.05

a) The null and alternative hypotheses are:

H0 : u = 28.5%

H1 : u < 28.5%

b) Given a 0.05 significance level.

This is a left tailed test.

The critical value =

$-Z_0.05 = -1.645$

The critical value = -1.645

c) Calculating the test statistic, we have:

$Z = \frac{p' - P}{\sqrt{\frac{P(1-P)}{n}}}$

$Z = \frac{0.2625 - 0.285}{\sqrt{\frac{28.5(1-28.5)}{800}}}$

Z = -1.41

d) Decision:

We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645

e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.

5 0

3 years ago