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crimeas [40]
2 years ago
8

I need help solving this difficult math problem step by step to get the answer. -10+-30/10 x 3

Mathematics
1 answer:
MAXImum [283]2 years ago
5 0

Answer:

The answer is -19

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-3(-x+4)=5+3(x+1) please help​
Lostsunrise [7]

Answer:

No solution.

Step-by-step explanation:

Step 1: Write equation

-3(-x + 4) = 5 + 3(x + 1)

Step 2: Solve for <em>x</em>

  1. Distribute: 3x - 12 = 5 + 3x + 3
  2. Combine like terms: 3x - 12 = 3x + 8
  3. Subtract 3x on both sides: -12 ≠ 8

Here, we see that <em>x</em> has to equal no solution. No value of <em>x</em> would make the equation true.

3 0
3 years ago
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I need this answer asap! Thanks, See attached image bellow. <br> 40+ points!
Wewaii [24]

Answer:

rectangle

Step-by-step explanation:

3 0
3 years ago
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!!!!!!!!50 POINTS. PLEASE HELP ASAP!!!!!!!
WARRIOR [948]

Answer:

x^6 + 2x^3  -2x^2 -8x + 4

Step-by-step explanation:

Okay, so we need to multiply f and g.

Since you said asap, I won't do the long explanation...

       x3        -4x       2

x2    x6       -4x2     2x2

2     2x3      -8x      4

x^6 + 2x^3  -2x^2 -8x + 4

3 0
3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
How much grain can this container hold?
ludmilkaskok [199]

Answer:

953.78 in^3 of grain.

Step-by-step explanation:

The volume of a cylinder = π r^2 h.

For this cylinder r =  radius of the base = 9/2

= 4.5 in  and the height h = 15 in

So the volume is 3.14 * 4.5^2 * 15

= 953.78 in^3.

8 0
3 years ago
Read 2 more answers
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