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Jet001 [13]
3 years ago
14

PLEASE help with these questions?? there are 7 of them! Thank you!!

Mathematics
1 answer:
svetlana [45]3 years ago
4 0
We know in these kind of triangles that have a 90 degrees angle, we can do:

{a}^{2} + {b}^{2} = {c}^{2}

so the first question:

\sqrt{ {99}^{2} + {20}^{2} } = 101

and then:

b = \sqrt{ {17}^{2} - {15}^{2} } = 8

so , when you want to rationalize , usually you must multiply the irrational factor as 1:

\frac{ \sqrt{15} }{ \sqrt{15} } = 1

so , now multiply:

\frac{3}{ \sqrt{15} } \times \frac{ \sqrt{15} }{ \sqrt{15} } = \frac{3 \sqrt{15} }{15} = \frac{ \sqrt{15} }{5}

in the next question,we know negative numbers in radical don't have any solution so 2 is correct,

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Annika borrowed a book from the library and forgot to return it on time, the library charges .25 per day for the first four days
Delicious77 [7]
So work it out as 0.25x4 =£1
Then 4-1=£3
£3 / 0.30 =10
10+4
So it is 14 days late
6 0
3 years ago
What is the gcf of 25 and 14
katrin2010 [14]
The gcf of 25 and 14 would be 2
3 0
3 years ago
Read 2 more answers
Which is an equation of the line containing the points (2,5) and (4,4) in standard form?
guapka [62]
First, find the slope of the line thru these 2 pts:

         5-4
m = --------- = -1/2
         2 -4 

Use the slope-intercept formula y = mx + b:

5 = (-1/2)(2) + b.  Then the y-intercept is 5 + 1, or 6:  y = (-1/2)x + 6

Multiplying this entire result (3 terms) by 2 results in 2y = -x + 12, or

x + 2y = 12 (answer)
6 0
3 years ago
I really need help please quickly!!<br> Will name brainliest!!
abruzzese [7]

Answer:

Q1)  7/8

Q2) 1/2

Q3) (-3)

Q4) 1

Step-by-step explanation:

(-2,-2) (-10,-9)

slope = (y₂ - y₁)/(x₂-x₁)

          =  (-9 -{-2}) / (-10-{-2})

           = (-9+2)/(-10+2) = -7/-8

          = 7/8

Q2) Slope= (-2+4)/7-3 = 2/4 = 1/2

Q3) slope = (4-10)/(-1+3) = -6/2 = (-3)

Q4) slope = (0+2)/(-1+3) = 2/2 = 1

5 0
3 years ago
Suppose the weights of seventh‑graders at a certain school vary according to a Normal distribution, with a mean of 100 pounds an
alex41 [277]

Answer:

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 100 pounds

Sample mean, \bar{x} = 98 pounds

Sample size, n = 35

Alpha, α = 0.05

Population standard deviation, σ = 7.5 pounds

First, we design the null and the alternate hypothesis

H_{0}: \mu = 100\text{ pounds}\\H_A: \mu < 100\text{ pounds}

We use one-tailed(left) z test to perform this hypothesis.

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{98 - 100}{\frac{7.5}{\sqrt{35}} } = -1.577

Now, z_{critical} \text{ at 0.05 level of significance } = -1.64

We calculate the p-value with the help of standard normal table.

P-value = 0.057398

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept is.

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.

3 0
3 years ago
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