Write the equation of the line that passes through the point (0,16) with a slope of 4/5 or 4 over 5

anyone know the answer i worked it out but didn't know where i messed up so i am just gonna ask for someone to help me with it.

hram777 [196]

I also worked it out but I got -5/3

2(6p-5) ≥ 3(p-8) -1

12p-10 ≥ 3p-24-1

12p-10 ≥ 3p-25

12p-3p ≥ -25+10

9p ≥ -15

-15/9

P ≥ -5/3

5 0

3 years ago

(no links you bot) name at least two independent variables that could result in a change in a monthly electric bill

viva [34]

Answer:

In the case of the monthly electricity cost, the independent variables could also include the non-production machines using electricity, the physical size of the products, the skill level of the operators, the outside temperature and humidity, etc.

Step-by-step explanation:

LOL

hope this helps

7 0

3 years ago

Read 2 more answers

X³ by a³ + a³by x³=p x³by a³- a³by ³=q eliminate x

sasho [114]

Answer:

Step-by-step explanation:

$\frac{x^3}{a^3} +\frac{a^3}{x^3} =p\\\frac{x^3}{a^3} -\frac{a^3}{x^3} =q\\adding\\2\frac{x^3}{a^3} =p+q\\\frac{x^3}{a^3} =\frac{p+q}{2} \\substitute~in~first\\\frac{p+q}{2} +\frac{2}{p+q} =p\\\frac{(p+q)^2+4}{2(p+q)} =p\\(p+q)^2+4=2p(p+q)\\p^2+q^2+2pq+4-2p^2-2pq=0\\q^2-p^2+4=0\\q^2=p^2-4$

6 0

3 years ago

The net best represents which of the solids?

Debora [2.8K]

Answer:

Where is the picture for the question?

5 0

3 years ago

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample x⁻ = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

( $(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )$

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{28-30 }{\frac{10}{\sqrt{25} } }$

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0

3 years ago