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3 years ago

Which equation can be used to find 150 percent of 500?

Mathematics

2 answers:

tiny-mole [99]3 years ago

8 0

Answer:

ITS DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDd

Step-by-step explanation:

grandymaker [24]3 years ago

4 0

Answer:

1.5 x 500=x

Step-by-step explanation:

We know that "percent" means "out of 100" and "of" means multiply. Using that knowledge, we can say 150/100 x 500 is the answer. 150/100 is 1.5, so the equation could also be 1.5 x 500=x

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Milton makes trail mix for his hiking group he makes a 1 and 1/4 pound of peanuts 14 oz of raisins 12 oz of walnuts in 10 oz of

strojnjashka [21]

Answer:

Each hiker receives 7 ounces of trail mix

Step-by-step explanation:

Trail mix=quantity of peanuts+quantity of raisins+quantity of walnuts+quantity of chocolate chips

where;

Quantity of peanuts=1.25 pounds

Quantity of raisins=14 oz, since 1 pound=16 oz.

Quantity of raisins=(14/16)=0.875 pounds

Quantity of walnuts=12 oz=(12/16)=0.75 pounds

Quantity of chocolate chips=10 oz=(10/16)=0.625 pounds

Replacing;

Trail mix=(1.25+0.875+0.75+0.625)=3.5 pounds

Trail mix=Quantity per hiker×Number of hikers

where;

Trail mix=3.5 pounds

Quantity per hiker=q

Number of hikers=8

Replacing;

3.5=q×8

q=3.5/8=0.4375 pounds

I pound=16 ounces

q=0.4375×16=7 ounces

Each hiker receives 7 ounces of trail mix

8 0

3 years ago

In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

<u>First</u> we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

<u>Next</u> we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$