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RideAnS [48]
2 years ago
12

What is -2/3 h +3< -9 ?

Mathematics
2 answers:
Nastasia [14]2 years ago
7 0

Answer:

18

Step-by-step explanation:

subtract 3 from both sides

then divide by -2/3 from both sides and flip the inequality sign

then you get h > 36/2

Then h > 18

klasskru [66]2 years ago
6 0

Answer: h>18

Step-by-step explanation:

variable  by dividing each side by factors that don’t contain the variable .

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What is the value of x in the equation 2x - 3 = 9- 4x? o 6 o 3 O 1 02​
riadik2000 [5.3K]

Answer:

2? The last one

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand
sesenic [268]

Answer:

The 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =80 represent the sample mean 1

\bar X_2 =75 represent the sample mean 2

n1=25 represent the sample 1 size  

n2=36 represent the sample 2 size  

\sigma_1 =5 sample standard deviation for sample 1

\sigma_2 =3 sample standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =80-75=5

Since the Confidence is 0.94 or 94%, the value of \alpha=0.06 and \alpha/2 =0.03, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that z_{\alpha/2}=1.88  

The standard error is given by the following formula:

SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118

Confidence interval

Now we have everything in order to replace into formula (1):  

5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898  

5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102  

So on this case the 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102  

7 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
A local charity is selling seats to a baseball game. Seats cost $20 each, and snacks cost an additional $5 each.
vlada-n [284]

Answer:

let seat as x,

and snack as y.

so the linear equation is 20x+5y=400.

7 0
2 years ago
I’m confused on this one
RoseWind [281]

It's reflection; it's not translation

so answer is the last one.

d. No; it is one reflection after another with respect to the two parallel lines.

hope it helps

7 0
3 years ago
Read 2 more answers
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