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3 years ago

What number would complete the pattern? 107 98 90 83 77 68 65

1 answer:

tatiyna3 years ago

8 0

The numbers are part of an arithmetic progression, where the difference between each succeeding number is reduced by 1 each time. Between 107 and 98, the difference is 9. Between 98 and 90, the difference is 8, and so on. This is shown below:

107 - 98 = 9
98 - 90 = 8
90 - 83 = 7
83 - 77 = 6
77 - x = 5
x - 68 = 4
68 - 65 = 3

From that, we can solve for x using either of the equations:

77 - x = 5
x = 77 - 5
x = 72

Therefore 72 would complete the pattern.

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Please help it will be brainlist as long as it’s right

Vikentia [17]

Answer:

x=32

Step-by-step explanation:

x+12=x+11+33

x+12=44

x=44-12

x=32

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3 years ago

Pachacha [2.7K]

Answer:

New area: 12ft^2

Step-by-step explanation:

1/3 of 9 is 3

1/3 of 12 is 4

Area = l x w

so 3 x 4 = 12

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3 years ago

Find the nearest number before and after any real number.

Mkey [24]

Answer:

add 1 and subtract 1

Step-by-step explanation:

so if u add and subtract 1. take 5. 5+1 and -1 is 4 and 6

Hope this helps :D

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3 years ago

Kate has 48 softballs. She wants to divide them evenly among b softball bags. Which expression represents how many softballs she

SOVA2 [1]

Answer:

Softballs = (48/b)

Step-by-step explanation:

Given that,

Total number of softballs = 48

She wants to divide them evenly among b softball bags.

We need to find an expression that represents how many softballs she should put into each bag.

As it is dividing among b softball, it means we divide 48 and b. So,

$s=\dfrac{48}{b}$

Hence, the required expression is (48/b).

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