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3 years ago

7

Select each expression that is equivalent to 3/16 if x = 3/4

1 answer:

iren2701 [21]3 years ago

3 0

Answer:

Step-by-step explanation:

B

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The first house on the street has a lawn that is 300 feet long. The second house

Ivanshal [37]

Answer: 450

Step-by-step explanation:

first house--300

second house--50

third house--50 x2=100

300+50+100=450

5 0

3 years ago

How to do? Can anyone help me please? I will appreciate that a lot!

Nitella [24]

Triangle ABD is i<span>nscribed in circle and </span>has hypotenuse as circle's diameter so angle ABD must be 90°

Since it is parallelogram, ∠ABD = ∠BDC = 90° and it is given that ∠ADB = 30°

So we have ∠CDO = ∠CDB + ∠ADB = 30° + 90° = 120°

Final answers: ∠CDO = 120° and ∠ABD = 90°

4 0

3 years ago

HURRY PLEASE HELPPPPPPP

elena-s [515]

Answer:

I think the answer is D

if we use this equation to find the slope ( slope= Y2-Y1/X2-X1)

6-3= 3

2-1=1

so then you divide it and you get 3/1 or 3

4 0

3 years ago

13+39-2-10-6 power2 orders of operation

Advocard [28]

6 to the power of 2 is 6 multiplied by six which is 36
then 13 + 39 is 52 - 2 = 50- 10 = 40 - 36= the answer 4

8 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