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2 years ago

12

15 points hurry please

1 answer:

gavmur [86]2 years ago

3 0

Answer:

the third and fifth one

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A banner is in the shape of a right triangle. the area is 63 inches. The height of the banner is 4 in less than twice the width

Blizzard [7]

Area=1/2 times base times height
note:bh=base times height

a=1/2bh
b=width

h=-4+2w
h=2w-4
subsitute
a=1/2w(2w-4)
a=1/2(2s^2-4w)
a=w^2-2w
a=63
63=w^2-2w
subtract 63 from both sdies
0=w^2-2w-63
factor
find what 2 numbers multiply to get -63 and add to get -2
the numbers are -9 and 7
so
0=(w-9)(w+7)
if xy=0 then x and/or y=0

so
w-9=0
w+7=0

solve each
w-9=0
add 9 to both sdies
w=9

w+7=0
subtract 7 from both sides
w=-7
width cannot be negative so this can be discarded

width=9

subsitute
l=2w-4
l=2(9)-4
l=18-4
l=14

legnth=14 in
width/base=9 in

5 0

2 years ago

A 10K run is 10 kilometers long. If 1 meter = 1.094 yards, use dimensional analysis to find the length of the race in miles. (Hi

laila [671]

10936.1 yards = 10 kilometers 1 kilometer = 1093.1 yards

7 0

3 years ago

Read 2 more answers

Mariah just read the following passage in a short story:

Nana76 [90]

Answer:

D. enlargement

Step-by-step explanation:

enlargement means to increase something

3 0

2 years ago

Read 2 more answers

Use the figure to find the radius.

BlackZzzverrR [31]

The answer is 4 (four)

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2 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