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2 years ago

Please help i will mark branliest

Mathematics

1 answer:

earnstyle [38]2 years ago

5 0

Answer:

Step-by-step explanation:

sorry if im wrong

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Jackie's restaurant bill was $23. 50, and the tax was 8%. Jackie left a 20% tip on the post-tax bill. How much was the tip Jacki

Ksju [112]

well, the subtotal or pre-tax bill is $23.50, now let's add the 8% tax and

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8\% of 23.50}}{\left( \cfrac{8}{100} \right)23.50}\implies 1.88$

so the post-tax bill is 23.50 + 1.88 = 25.38, and Jackie's tip is 20% of that

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{20\% of 25.38}}{\left( \cfrac{20}{100} \right)25.38}\implies 5.076\qquad \approx 5.08$

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1 year ago

Classify the model as Exponential GROWTH or DECAY.<br> A=1200(.85)6

BartSMP [9]

Answer: DECAY

Step-by-step explanation:

(.85) is less than 1

3 0

2 years ago

The two triangles formed by the flagpole and the man shown below are similar.

vagabundo [1.1K]

It’s gonna be 51 feet

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2 years ago

Read 2 more answers

Sandra is cooking her famous brownies for her class. For her recipe, she needs 5/7 of a cup of sugar, and 1/8 as much salt as su

jarptica [38.1K]

Answer:

$\frac{5}{56}$ salt does she need for her recipe.

Step-by-step explanation:

As per the statement:

For her recipe:

She needs = $\frac{5}{7}$ of a cup of sugar.

It is also given that: 1/8 as much salt as sugar.

⇒ She needs salt for her recipe = $\frac{1}{8} \cdot \frac{5}{7} = \frac{5}{7 \cdot 8} = \frac{5}{56}$

Therefore, $\frac{5}{56}$ salt does she need for her recipe.

7 0

3 years ago

According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif

suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

$\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018$

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

$z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55$

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

$P(0.79$

3 0

3 years ago