Who has insta gramm I really need help with my assignments I need good grades to graduate

Tell me a fun fact about horses will give BRAINLIEST to whoever tells me something I don’t know (I have been riding for 12 years

Nonamiya [84]

Here are a few fun facts:

A 19th century horse named 'Old Billy' is said to have lived 62 years.

Horses can sleep both lying down and standing up.

Horses can run shortly after birth.

Horses have around 205 bones in their skeleton.

Horses have been domesticated for over 5000 years.

Horses use their ears, eyes and nostrils to express their mood.

Because horse’s eyes are on the side of their head they are capable of seeing nearly 360 degrees at one time.

The fastest recorded sprinting speed of a horse was 88 kph (55 mph). Most gallop at around 44 kph or 27 mph.

The Przewalski’s horse is the only truly wild horse species still in existence. The only wild population is in Mongolia. There are however numerous populations across the world of feral horses e.g. mustangs in North America.

I hope you learned something new!

4 0

2 years ago

Steve and Michelle Neely are celebrating their 30th anniversary by having a reception at a local reception hall. They have budge

lord [1]

4 500-60 = 4 440
4 440/38 = 116.8
At most they would be able to invite 116 people to the reception with their budget.

8 0

3 years ago

Read 2 more answers

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago

PLZ HELP ASAP I WILL GIVE BRAINIEST TO FIRST PERSON TO ANSWER. question: Tahmar knows the formula for simple interest is I = Prt

Gre4nikov [31]

P=I/RT
P=20/(0.05*5)
P=20/(0.25)
P=80

It is $80

3 0

3 years ago

Read 2 more answers

Answer for both of them<br> 3(4x - 1)<br> 2x + 4

kobusy [5.1K]