All categories

3 years ago

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

You might be interested in

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, $\sigma=8.8$ g

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°

Alternative working:

∠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