Which point is located at -0.905?

Solve the following exponential equations. (10^x)^2 − 2(10^x) − 1 = 0

Hunter-Best [27]

Answer:

x = log(1+√2) ≈ 0.38277569

Step-by-step explanation:

Let z = 10^x. Then the equation can be rewritten ...

z^2 -2z -1 = 0

z^2 -2z +1 -2 = 0 . . . . rewrite to a form with a square

(z -1)^2 = 2 . . . . . . . . . add 2

z -1 = ±√2 . . . . . . . . . . square root

z = 1 ± √2 . . . . . . . . . . add 1

From the definition of z, we know it must be positive, so the only viable solution here is ...

z = 10^x = 1+√2

x = log(1+√2)

x ≈ 0.38277569

6 0

3 years ago

Could someone help me multiply by 1-Digit numbers? Could you help me with 31×7 Using that ^

Alja [10]

Answer:

the answer would be 31^7= 217

Step-by-step explanation:

hope this helps!

4 0

4 years ago

To build a desktop you put 2 boards together . One board is 1/3 inch thick, and the other is 1/2 inch thick . How thick will the

likoan [24]

The answer is 5/6 inches thick

3 0

3 years ago

Find the product Give your answer in simplest form. 3/5x12

Alex

The origainal answer is 36/5 but the simplest form is 7 1/5

4 0

3 years ago

Read 2 more answers

A student at a four-year college claims that average enrollment at four-year colleges is higher than at two-year colleges in the

Kitty [74]

Answer:

a) Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

b) $t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

c) $p_v =P(t_{68}>0.0921)=0.463$

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

Step-by-step explanation:

Information given

$\bar X_{1}=5069$ represent the mean for two year colleges

$\bar X_{2}=5216$ represent the mean for four year college

$s_{1}=4773$ represent the sample standard deviation for 1

$s_{2}=8141$ represent the sample standard deviation for 2

$n_{1}=35$ sample size for the group 2

$n_{2}=35$ sample size for the group 2

$\alpha=0.05$ Significance level provided

Part a

Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

Part b

We want to test if the mean of enrollment for four year college is higher than for two year college, the system of hypothesis would be:

Null hypothesis: $\mu_{2}-\mu_{1} \leq 0$

Alternative hypothesis: $\mu_{2} - \mu_{1} > 0$

The statistic is given by:

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

The degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$ Replacing the info given we got:

$t=\frac{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921$

Part c

The p value would be given by:

$p_v =P(t_{68}>0.0921)=0.463$

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

6 0

4 years ago