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3 years ago

In the equation 9 more than 3 times a number is -6, what does a number equal

1 answer:

8 0

So 9 more than 3 times a number, would be:
9+(3x)=-6
then you would solve
subtract 9 from 9 and 9 from -6
you get 3x=-15
divide by 3 on both sides
x=-5

your welcome!!!! :)

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Please Help!!!

Viefleur [7K]

Answer:

Numeric expressions apply operations to numbers. For example, 2(3 + 8) is a numeric expression. Algebraic expressions include at least one variable and at least one operation (addition, subtraction, multiplication, division). For example, 2(x + 8y) is an algebraic expression.

Step-by-step explanation:

8 0

3 years ago

In which direction does the parabola open?

svetoff [14.1K]

The parabola opens up as the largest term has a positive leading co-efficent

7 0

2 years ago

-3 + 5n > -13 <br> i need help on this someone please help

Butoxors [25]

Answer:

n > -2

Step-by-step explanation:

5n > -10

n > -2

8 0

2 years ago

Read 2 more answers

Choose the correct simplification of the expression (2x2y6z5)(5x4y5z3).

erastovalidia [21]

Answer:

The correct option is 4.

Step-by-step explanation:

The given expression is

$(2x^2y^6z^5)(5x^4y^5z^3)$

Simplify the given expression.

$(2x^2y^6z^5)\times (5x^4y^5z^3)$

$(2\times 5)(x^2x^4)(y^6y^5)(z^5z^3)$

$10x^{2+4}y^{6+5}z^{5+3}$ $[\because x^mx^n=x^{m+n}]$

$10x^6y^{11}z^8$

Therefore correct option is 4.

8 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago