Please answer quickly and explain how!ToT<br><br><br> 276 is what percent of 92?

Every week malik spends 5 2/15 hours doing his math homework and 2 13/30 hours doing his science homework. Estimate how much mor

ArbitrLikvidat [17]

This is Simple, Just Subtract Time taken by Science homework from time taken by Maths homework..

$= 5 \frac{2}{15} - 2 \frac{13}{30} \\ = \frac{150}{15} - \frac{73}{30} \\ = \frac{300 - 73}{30} \\ = \frac{227}{30} \\ = 7 \frac{17}{30} \: hours \: more.$

!! Hope It Helps !!

3 0

4 years ago

What is the inverse of f(x)=4x

Allisa [31]

The sum can be rewritten as y=4x, where y=f(x).
Now, we can rewrite the equation a x=y/4
Therefore, inv(f(x))=x/4

3 0

4 years ago

Read 2 more answers

Plz help I'm stuck on this one

Zarrin [17]

C. 12 Hope this helps!

8 0

4 years ago

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Correct answers only please!

Alla [95]

Answer:

$15865.6

Step-by-step explanation:

Using the given formula :

I = P × R × T

I = 99.160 × 8/100 × 2

I = $ 15865.6

6 0

3 years ago

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The germination rate is the rate at which plants begin to grow after the seed is planted.

chubhunter [2.5K]

Answer:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

$p_v =P(z$

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, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

$\hat p=\frac{7}{15}=0.467$ estimated proportion of seeds germinated

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

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 germinated seeds is less than 0.9 or 90%.:

Null hypothesis: $p\geq 0.9$

Alternative hypothesis: $p < 0.9$

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:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

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, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

4 0

3 years ago

Please answer quickly and explain how!ToT 276 is what percent of 92?