All categories

2 years ago

What is the value of x ? 7x– 5 = 4x - 14

Mathematics

1 answer:

ruslelena [56]2 years ago

7 0

Answer: X is equal to -3

Step-by-step explanation: I isolated x and then divided both sides by x

You might be interested in

5 + (-3) is units from 5 in which direction

Ymorist [56]

Answer:

Step-by-step explanation:

Starting at 5 on the number line, 5 + (-3) is 3 units to the left of 5; that is, you are at 2 on the number line.

4 0

3 years ago

A scale on a map shows that 2.52.52, point, 5 centimeters represents 151515 kilometers.

Dmitry_Shevchenko [17]

Answer:

105 km

Step-by-step explanation:

Written as a proportion, the ratio of km to cm will be the same for both distances:

(x km)/(17.5 cm) = (15 km)/(2.5 cm)

x km = (15 km)(17.5/2.5) = 105 km

105 actual kilometers are represented by 17.5 cm on the map.

5 0

3 years ago

2. What is the slope passing through the points (2, -1) and (-4, 11)

algol [13]

Answer:

I believe it is y equals 3 minus 2x.

4 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Find the missing length

bonufazy [111]

Apply the Pyth. Theorem:

c^2 = 3^2 + 7^2, or c^2 = 9 + 49 = 58. Thus, c = sqrt(58).

7 0

3 years ago