Please help me!<br> [One Step Inequalities]

Solve the equation. Round to the nearest thousandth. -12e-x + 8 = 7

Marysya12 [62]

Answer:

-12e-x + 8 = 7

-12e-x = 7-8

-12e-x = -1

12e-x = 1

e-x = 1/12

ln(e-x) = ln(1/12)

x = -ln(1/12)

x = -2.485

7 0

2 years ago

Suppose your friends parents invest $15,000 in an account paying 6% compounded annually. What will be the balance after 9 years

JulijaS [17]

A = 15000(1.04)6 .........You can put this into your calculator such that it becomes
A = $18,979.79

6 0

2 years ago

How to increase 42kg by 3%

lbvjy [14]

Multiply .03 by 42

.03×42=1.26

1.26kg + 42kg= 43.26

Answer: 43.26kg

4 0

3 years ago

Dominic invested $520 in an account in the year 1995, and the value has been growing exponentially at a constant rate. The value

Tom [10]

Answer:

I believe the answer is $936.64

Step-by-step explanation:

I hope this helps!

6 0

3 years ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2>0$

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

3 years ago

Please help me! [One Step Inequalities]