Answer:
6xrx7
2
Regroup terms.
6x{x}^{7}r6xx7r
3
Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}xaxb=xa+b.
6{x}^{1+7}r6x1+7r
4
Simplify 1+71+7 to 88.
6{x}^{8}r6x8r
Answer:
The fraction of chores which Janie wants to earn enough money to buy a CD for $13.50 are 8.75~9 chores
Step-by-step explanation:
Given that Janie has $3. She earns $1.20 for each chore she does and can do fractions of chores. She wants to earn enough money to buy a CD for $13.50.
we have to find the fraction of chores Janie did to earn total of $13.50
Let the fraction of chores is x which Janie wants to earn enough money to buy a CD for $13.50.
Total money=$3+$1.20(fraction of chores)
$13.50=$3+$1.20(x)
$13.50-$3=$1.20(x)
x=8.75~9
Hence, the fraction of chores which Janie wants to earn enough money to buy a CD for $13.50 are 8.75~9 chores
The graph for 2x-3y=12 is
Answer:
a) Alternative hypothesis should be one sided. Because Null and Alternative hypotheses are:
: μ=2.66 dyne-cm.
: μ<2.66 dyne-cm.
b) the hypothesis that mean adhesion is at least 2.66 dyne-cm is true
Step-by-step explanation:
Let μ be the mean adhesion in dyne-cm.
a)
Null and alternative hypotheses are:
: μ=2.66 dyne-cm.
: μ<2.66 dyne-cm.
b)
First we need to calculate test statistic and then the p-value of it.
test statistic of sample mean can be calculated as follows:
t=
where
- M is the mean adhesion assumed under null hypothesis (2.66 dyne-cm)
- s is the standard deviation known (0.7 dyne-cm_2)
Sample mean is the average of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm, that is
≈ 3.37
using the numbers we get
t=
≈ 2.27
The p-value is ≈ 0.043. Taking significance level as 0.05, we can conlude that sample proportion is significantly higher than 2.66 dyne-cm.
Thus, according to the sample the hypothesis that mean adhesion is at least 2.66 dyne-cm is true