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den301095 [7]
3 years ago
15

NEED HELP ASAP!!!!!!! WILL GIVE BRAINLIEST!!!!!!

Mathematics
1 answer:
sattari [20]3 years ago
6 0
I belive it D or C srry if im wrong :( ill give u it back
You might be interested in
A Food Marketing Institute found that 39% of households spend more than $125 a week on groceries. Assume the population proporti
Anuta_ua [19.1K]

Answer:

0.6210

Step-by-step explanation:

Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries

Sample size n =87

Sample proportion will follow a normal distribution with p =0.39

and standard error = \sqrt{\frac{0.39(1-0.39)}{87} } \\=0.0523

the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

=P(0.29

There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

7 0
4 years ago
Suppose that the store manager of a small rural pharmacy is doing a linear regression of daily sales of over-the-counter (OTC) d
Doss [256]

Answer:

3.83

Step-by-step explanation:

Mean of x = Σx / n

Mean of x = (14 + 19 + 13 + 6 + 9) / 5 = 12.2

Sum of square (SS) :

(14-12.2)^2 + (19-12.2)^2 + (13-12.2)^2 + (6-12.2)^2 + (9-12.2)^2 = 98.8

Mean of y = Σy / n

Mean of y = (101 + 89 + 48 + 21 + 47) / 5 = 61.2

Σ(y - ybar)² = (101-61.2)^2 + (89-61.2)^2 + (48-61.2)^2 + (21-61.2)^2 + (47-61.2)^2 = 4348.8

df = n - 2 = 5 - 2 = 3

Σ(y - ybar)² / df = 4348.8 / 3 = 1449.6

√(Σ(y - ybar)² / df) = √1449.6 = 38.074

Standard Error = √(Σ(y - ybar)² / df) / √SS

Standard Error = 38.074 / √98.8

Standard Error = 3.83

4 0
3 years ago
CAN SOMEONE HELP ME WITH THISSSS
Semmy [17]

Answer:

no i cannot

Step-by-step explanation:

4 0
3 years ago
L has $200. Ty has 30% more than Lu and twice as much as Ali. How much money do they have altogether? (PLEASE SHOW STEPS!)
saul85 [17]
30% of 200=60 200+60= 260 
Ty has £260
£260 x 2= £520

200+260+520= £980

Answer £980

I hope this is right :D
5 0
4 years ago
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
4 years ago
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