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2 years ago

10

Jim earns $1,600 per month after taxes. He is working on his budget and had the first three categories finished. Why will he hav

e a problem with the rest of his budget?

1 answer:

lukranit [14]2 years ago

3 0

B.

He has used the highest recommended percentages to calculate the amounts for the three categories.

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I need help on this it is a fraction subtract another fraction which is negative thx to whoever helps 5/7-(-1/7)

Sav [38]

Answer:

$\frac{5}{7}-(-\frac{1}{7})=\frac{6}{7}$

Step-by-step explanation:

To evaluate :

$\frac{5}{7}-(-\frac{1}{7})$

Solution:

Two negatives multiply to become a positive.

Thus, we can remove parenthesis by reversing the signs of the fraction by multiplying the negative outside.

⇒ $\frac{5}{7}+\frac{1}{7}$

Since the denominators are same for both fractions, so we simply add the numerators.

⇒ $\frac{5+1}{7}$

⇒ $\frac{6}{7}$ (Answer)

6 0

2 years ago

F(x) = 5x^3+ 2x^2 - 90x - 36. <br><br> finding all zeros by factoring and explaining the steps

ella [17]

Answer:

f(x)=5x^3-2x^2-90x-36=0

=x^2(5x-2)-18(5x-2)=(x^2-18)(5x-2)=0

x^2-18=0/5x-2=0

x^2=18=x=9√2

5x-2=0

x=2/5

zeros are 9√2,2/5

8 0

2 years ago

If a turtle can move 1/7 of a km every hour, then how many hours would it take for a turtle to go 5 km?

Paul [167]

Answer:

it take 1/7 km in 1 hour then,

1 km it take 7 hours

so, it cover 5 km in 35 hours

1/7 km ------- 1 hour

1km ------- 7 hour

5km -------( 5*7) == 35 hours

5 0

2 years ago

What is the factored form of x^3-1

polet [3.4K]

Answer:

the factors of x^3-1 are $x^3 -1 = (x-1) (x^2 +x+1)\\$

Step-by-step explanation:

We need to find factors of x^3 -1

$x^3-1 \\(x)^3 -(1)^3\\We \,\, know\,\, a^3 -b^3 = (a-b) (a^2 +ab + y^2) \\So, \,\, using\,\, the \,\, formula\,\,\\Here\,\, a= x \,\,and\,\, b= 1\\(x)^3 -(1)^3 = (x-1) (x^2 +x+1)\\$

So, the factors of x^3-1 are $x^3 -1 = (x-1) (x^2 +x+1)\\$

7 0

3 years ago

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two bakeries' specialty pi

DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ <em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ <em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0

3 years ago