Answer:
i hope this helps
Step-by-step explanation:
345 mi; if you travel 69 miles per hour, and drive 5 hours, you have to multiply 69 and 5 to find how far you travel in 5-- 69 times 5 is 345.
Answer:
No, because the 95% confidence interval contains the hypothesized value of zero.
Step-by-step explanation:
Hello!
You have the information regarding two calcium supplements.
X₁: Calcium content of supplement 1
n₁= 12
X[bar]₁= 1000mg
S₁= 23 mg
X₂: Calcium content of supplement 2
n₂= 15
X[bar]₂= 1016mg
S₂= 24mg
It is known that X₁~N(μ₁; σ²₁), X₂~N(μ₂;δ²₂) and σ²₁=δ²₂=?
The claim is that both supplements have the same average calcium content:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
The confidence level and significance level are to be complementary, so if 1 - α: 0.95 then α:0.05
since these are two independent samples from normal populations and the population variances are equal, you have to use a pooled variance t-test to construct the interval:
[(X[bar]₁-X[bar]₂) ±
*
]


[(1000-1016)±2.060*23.57*
]
[-34.80;2.80] mg
The 95% CI contains the value under the null hypothesis: "zero", so the decision is to not reject the null hypothesis. Then using a 5% significance level you can conclude that there is no difference between the average calcium content of supplements 1 and 2.
I hope it helps!
<h3>
Answer: 5/162</h3>
===================================================
Work Shown:
A = probability of landing on an odd number
A = 5/9 because there are 5 odd numbers out of 9 total
B = probability of getting heads
B = 1/2
C = landing on 3 on the spinner
C = 1/9
Multiply the values of A,B,C
A*B*C = (5/9)*(1/2)*(1/9) = (5*1*1)/(9*2*9) = 5/162
Note: The fraction 5/162 converts to the approximate decimal 0.03086 which then converts to 3.086%, when rounding to the nearest tenth of a percent we get 3.1%
Answer:he will use 45 minutes to burn 540 calories . he spend a total of 50 minutes ie including 5 minutes of rest.
Step-by-step explanation: 1 minute burns 12 calories
Therefore 540÷12 = 45 minutes plus 5 minutes of rest = 50 minutes.