Ask question

Ask question

All categories

photoshop1234 [79]

1 year ago

13

Please help me!!!!!!

1 answer:

Setler79 [48]1 year ago

7 0

Answer:

It is an obtuse triangle I think because

You might be interested in

Solve the quadratic equations below using the factorization method.

liubo4ka [24]

Answer: omg this is hard

Step-by-step explanation:

4 0

2 years ago

Si en 11 cajas de chocolates hay 220 chocolates en total. Calcula cuántos chocolates habrá en total en 5 cajas de chocolates.

pochemuha

Answer:

Habrá 100 chocolates en total en 5 cajas de chocolates.

Step-by-step explanation:

Esta pregunta es resolvida usando proporciones, con lo uso de regla de tres .

11 cajas, 220 chocolates:

Entonces:

$\frac{220}{11} = 20$

20 chocolates por caja.

Calcula cuántos chocolates habrá en total en 5 cajas de chocolates.

20 chocolates por caja, entonces:

5*20 = 100

Habrá 100 chocolates en total en 5 cajas de chocolates.

6 0

2 years ago

A 5-meter ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 0.4 m

Flura [38]

<h2>The top of the ladder is descending at 0.3 m/s.</h2>

Step-by-step explanation:

By Pythagoras theorem we know that

Hypotenuse² = Base² + Perpendicular²

h² = b² + p²

We have for ladder

h = 5 m

b = 3 m

5² = 3² + p²

p = 4 m

$\frac{db}{dt}=0.4m/s\\\\\frac{dh}{dt}=0$

Differentiating h² = b² + p² with respect to time

$2h\times \frac{dh}{dt}=2b\times \frac{db}{dt}+2p\times \frac{dp}{dt}\\\\5\times 0=3\times 0.4+4\times \frac{dp}{dt}\\\\\frac{dp}{dt}=-0.3m/s$

The top of the ladder is descending at 0.3 m/s.

3 0

2 years ago

20 POINTS !!! ONLY HELP IF UR 100% SURE !!!

Arte-miy333 [17]

Your answer is C for sure I am!

8 0

2 years ago

Read 2 more answers

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

a

$SE = 0.66}$

b

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago