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sp2606 [1]
3 years ago
13

Tim recorded that 27 out of 63 customers used a shoping cart at the grocery store.Based on these results, about how many custome

rs out of 350 can be expected to use a shoping cart? A) 110
B) 260
C) 90
D) 150
Mathematics
1 answer:
anygoal [31]3 years ago
7 0
I am estimating it is A
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What is 15.75 rounded to the whole number​
victus00 [196]

Answer:

C. 16

Step-by-step explanation:

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"A cylindrical canister contains 3 tennis balls. Its height is 8.75inches, and its radius is 1.5. The diameter of one tennis bal
puteri [66]
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Find Volume of 1 tennis ball:
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Volume of 1 tennis ball = 4/3 x 3.14 x (2.5 ÷ 2)³

Volume of 1 tennis ball = 8.18 in³

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Find Volume of 3 tennis balls:
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Volume of 3 tennis balls = 8.18 x 3 = 24.54 in³

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Find Volume of teh cylindrical canister:
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Volume of the cylindrical canister = 52.99 in³

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8 0
3 years ago
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-x=-15<br><br>Syep by step solution need​
andrew-mc [135]

Answer:

x = 15

Step-by-step explanation:

-x = -15

x = -1 × (-15)

<u>x</u><u> </u><u>=</u><u> </u><u>1</u><u>5</u>

6 0
2 years ago
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Please help! Will mark brainliest! 20 points!
notsponge [240]

8      3

    -

9      4


Get common denominators.

8(4)      3 (9)

        -

9(4)      4 (9)

=

32      27

      -

36      36

=

5/36

5 0
3 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
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