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Keith_Richards [23]
3 years ago
5

Can someone answer this?

Mathematics
1 answer:
Reil [10]3 years ago
6 0

Answer:

I got you it

Step-by-step explanation:

B and have a good day

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use the image below to describe at least three different ratios written in simplest form. Include at least one part to part rati
oksano4ka [1.4K]

Answer:

charts huh

Step-by-step explanation:

678/980=x1280 hope this helps

8 0
2 years ago
Read 2 more answers
An organic farm has been growing an heirloom variety of summer squash. A sample of the weights of 40 summer squash revealed that
natta225 [31]

Answer:

b. 0.0228

Step-by-step explanation:

We are given that

n=40

Mean,\mu=402.7 g

Standard deviation, \sigma=8.8g

We have to find the probability hat the mean weight for a sample of 40 summer squash exceeds 405.5 grams.

P(x>405.5)=P(\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{405.5-402.7}{\frac{8.8}{\sqrt{40}}})

P(x>405.5)=P(Z>\frac{2.8}{\frac{8.8}{\sqrt{40}}})

P(x>405.5)=P(Z>2.01)

P(x>405.5)=1-P(Z\leq 2.01)

P(x>405.5)=1-0.977784

P(x>405.5)=0.022216

Hence, option b is correct.

5 0
2 years ago
Find the value of x.
konstantin123 [22]

Answer:

x= 37.5°

Step-by-step explanation:

∠CBD

= 180° -75° (adj. ∠s on a str. line)

= 105°

∠BCD= ∠BDC (base ∠s of isos. △BCD)

∠BCD= x

∠BCD +∠BDC +∠CBD= 180° (∠ sum of △BCD)

x +x +105°= 180°

2x= 180° -105°

2x= 75°

x= 37.5°

<u>Alternative</u><u> </u><u>working</u><u>:</u>

∠BDA= (180° -75°) ÷2 (base ∠s of isos. △ABD)

∠BDA= 52.5°

∠BDA +∠BDC= 90°

52.5° +x= 90°

x= 90° -52.5°

x= 37.5°

6 0
3 years ago
Find the distance between (0, 0) and (3, 4).
Orlov [11]

Answer:

L²=(3-0)²+(4-0)²

L²=9+16

L²=25

L=5units

5 0
2 years ago
Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.
skad [1K]
DfLet f(x)=\sec x and g(x)=\sqrt x. Then

y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)

By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}

where u=g(x)=\sqrt x, so that y=f(g(x))=f(u)=\sec u. We have

\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}
\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u

and so

\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}
5 0
3 years ago
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