First we need to write the null and alternate hypothesis for this case.
Let x be the average number of text message sent. Then
Null hypothesis: x = 100
Alternate hypothesis: x > 100
The p value is 0.0853
If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.
If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.
If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.
Complete Question
Riya is applying to her garden. She applies it at a rate of 25, 000 cm³ of mulch for every m² of garden space. At what rate is Riya applying mulch in m³/m²
Answer:
0.25m³/m²
Step-by-step explanation:
We are told the Riya sprays mulch at 250,000cm³ per m²
To find the rate at which Riya is spaying the mulch in m³/m² we would have to convert 250,00cm³/m² to m³/m²
1 cm³ = 1 × 10^-6m³
250,000 cm³ = x m³
Cross Multiply
1 cm³ × xm³ = 250,000cm³ × 1 × 10^-6 m³
X
x m³ = 250,000cm³ × 1 × 10^-6 m³/1 cm³
= 0.25m³
Therefore, the rate at which Riya is spraying the mulch = 0.25m³/m²
Answer:
wat
Step-by-step explanation:
Answer:
Quarters = 20
Dimes = 14
Step-by-step explanation:
Dime = x = 10 cent = $0.1
Quarter = y = 25 cent = $0.25
x + y = 34 - - - (1)
0.1x + 0.25y = 6.40 - - - - (2)
From (1)
x = 34 - y
0.1(34 - y) + 0.25y = 6.40
3.4 - 0.1y + 0.25y = 6.40
3.4 + 0.15y = 6.40
0.15y = 6.40 - 3.40
0.15y = 3
y = 3 / 0.15
y = 20
x = 34 - y
x = 34 - 20
x = 14
Quarters = 20
Dimes = 14
