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ycow [4]
3 years ago
7

What is the slope of 3x - 5y = 10

Mathematics
2 answers:
Ket [755]3 years ago
5 0
The slope of, 3x-5y=10 would be m=3/5
SCORPION-xisa [38]3 years ago
5 0
3x-5y=10
-5y=-3x+10
5y=3x-10
y=3/5x-10/5
y=3/5x-2

slope is 3/5
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What is the range ? JUST AN ANSWER​
mash [69]

Answer:

16

Step-by-step explanation:

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Can someone help me with these 3 questions of you can and show the work too
Vesnalui [34]
Hi
1. Temp was 5 at noon and dropped 59 degrees at midnight. 5+(-59)= -54 (B)

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3 years ago
I need to know how to do the whole thing and understand it.
patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33

Variance:

s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

H_0:\;\mu_A=\mu_B

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

H_a:\;\mu_A>\mu_B

Test statistic (t):

The test statistic of a two-sample t-test is given by

t=\frac{\bar{x}_A-\bar{x}_B}{s_p}

Where sp is the pooled standard deviation given by

\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}t=\frac{2-3.33}{0.763}=-1.74

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

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