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likoan [24]
3 years ago
14

PLEASEE HELP, I’M CONFUSED!

Mathematics
2 answers:
hichkok12 [17]3 years ago
7 0

Answer:

200

Step-by-step explanation:

if x is the original number:

0.7x = 140\\7x = 1400\\x = 200

quester [9]3 years ago
3 0

Answer:

182

Step-by-step explanation:

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Can someone help please​
Andreyy89

Answer:

option a

Step-by-step explanation:

in figure 1: 4n+8 = 4×1=4+8=12

figure 2:4n+8 =4×2=8+8=16 and so on

8 0
3 years ago
What is 4(2×-7)+3(3×-1)=3(2-×)​
Mama L [17]

Answer:

x=37/20

Step-by-step explanation:

4(2x-7)+3(3x-1)=3(2-x)

8x-28+9x-3=6-3x

17x+3x=6+31

20x=37

8 0
3 years ago
Read 2 more answers
Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne
Pavlova-9 [17]

Answer:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

p_v =P(t_{(15)}>1.507)=0.076

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

\bar X_{CTS}=2.35 represent the mean for the sample CTS

\bar X_{N}=1.83 represent the mean for the sample Normal

s_{CTS}=0.88 represent the sample standard deviation for the sample of CTS

s_{N}=0.54 represent the sample standard deviation for the sample of Normal

n_{CTS}=10 sample size selected for the CTS

n_{N}=7 sample size selected for the Normal

\alpha=0.01 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis:\mu_{CTS} \leq \mu_{N}

Alternative hypothesis:\mu_{CTS} > \mu_{N}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{CTS}+n_{N}-2=10+7-2=15

Since is a one side right tailed test the p value would be:

p_v =P(t_{(15)}>1.507)=0.076

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0
3 years ago
The points (1, 3) and (−3, 7) lie on a line. where does the line cross the x-axis?
Anastasy [175]
First, we find the equation of the line...

(1,3),(-3,7)
slope = (7 - 3) / (-3 - 1) = 4/-4 = -1

y = mx + b
slope(m) = -1
use either of ur points.... (1,3)...x = 1 and y = 3
now sub into the formula and find b, the y int
3 = -1(1) + b
3 = -1 + b
3 + 1 = b
4 = b
so the equation for this line is : y = -1x + 4 which is usually written as :
 y = -x + 4

Now...to find where the line crosses the x axis (or the x intercept)...we sub in 0 for y and solve for x

y = -x + 4
0 = -x + 4
x = 4... so ur x intercept (where the line crosses the x axis) is : (4,0)

3 0
3 years ago
Find all solutions of the equation in the interval [0, 2pi).
mina [271]

Answer:

x = 0 , π

Step-by-step explanation:

-4 \sin x = 1 - cos^2 x

  • Rewrite it by using the identity \sin^2x + \cos^2x = 1

=> -4\sin x = \sin^2x

  • Add 4sin x to both the sides.

=> -4\sin x + 4\sin x = sin^2x + 4\sin x

=> \sin^2x + 4\sin x = 0

  • Take sin x common from the expression in L.H.S.

=> \sin x(\sin x + 4)=0

Here , we can get two more equations to find x.

1) \sin x(\sin x + 4)=0

  • Divide both the sides by sin x

=> \frac{\sin x(\sin x + 4)}{\sin x} = \frac{0}{\sin x}

=> \sin x + 4 = 0

  • Substract 4 from both the sides.

=> \sin x + 4 - 4 = 0 - 4

=> \sin x = -4

=> x = No \; Solution

2) \sin x(\sin x + 4)=0

  • Divide both the sides by (sin x + 4)

=> \frac{\sin x(\sin x + 4)}{\sin x + 4} = \frac{0}{\sin x + 4}

=> \sin x =  0

=> x = 0 \; , \pi over interval [0 , 2π).

5 0
3 years ago
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