PLEASE EXPLAIN: ASAP Express your answer in scientific notation 2.8*10^-3-0.00065=

Chris saved x dollars and spent half of his savings buying video games. After earning an additional $15 cutting grass, he had $5

olga55 [171]

Chris started with $70 in the savings account

What is the equation representing the scenario?

The equation representing the scenario painted in the question is: one half times x plus 15 equals 50

1/2*X+15=50

X/2+15=50

X/2=50-15

X/2=35

X=35*2

X=70

In short the amount Chris started with in the savings account is $70

Find out more about savings plan equation on:brainly.com/question/23630456

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4 0

1 year ago

5(x+2/3)=3x-2/3The answer is-2.Please show work so I can understand how to solve.

kramer

5(x + 2/3) = 3x - 2/3.....distribute the 5 through the parenthesis
5x + 10/3 = 3x - 2/3 ....subtract 3x from both sides
5x - 3x + 10/3 = 3x - 3x - 2/3...combine like terms
2x + 10/3 = - 2/3 ....subtract 10/3 from both sides
2x + 10/3 - 10/3 = -2/3 - 10/3...simplify
2x = - 12/3....reduce
2x = - 4....divide both sides by 2
x = -4/2....reduces to -2

or this way...
5(x + 2/3) = 3x - 2/3.....distribute
5x + 10/3 = 3x - 2/3....multiply by lowest common denominator of 3
15x + 10 = 9x - 2 ....subtract 9x from both sides
15x - 9x + 10 = -2 ...subtract 10 from both sides
6x + 10 - 10 = -2 - 10...simplify
6x = -12.....divide both sides by 6
x = -12/6 reduces to -2

6 0

3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago

What prime factorization 32 is between 1 and 10

olga55 [171]

Answer:

2 x 2 x 2 x 2 x 2, which can also be written 32 = 2⁵

Step-by-step explanation:

8 0

3 years ago

Which expression is equivalent to 4(9 + 7)? 36 + 7 13 + 11 36 + 28 13 + 7

Misha Larkins [42]

36+28 because 4(9+7) = 64 and to get this answer you do 4*9 + 4*7 = 36 + 28 = 64