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Leya [2.2K]
2 years ago
5

Find the Balance:

Mathematics
1 answer:
lubasha [3.4K]2 years ago
3 0

Answer:

branliy app is very bad and poor,

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Last week, the price of cherries at the corner deli was $4.99 per pound. This week, cherries at the same deli are on sale at a 1
sergij07 [2.7K]

Answer:

The price for 3.75 pounds of cherries is $17.775.

Step-by-step explanation:

The price las week was $ 4.99 and it got a discount of 15% for this week, this means that the price for this week is 100% of the one prior minus 15%, therefore it is 85% of the last price. We need to calculate the price per pound with this discount as shown below:

\text{this week} = \text{last week}*\frac{95}{100}\\\\\text{this week} = 4.99*0.95 = 4.74 \text{ per pound}\\

Since the price is $4.74 per pound and we want to buy 3.75 pounds, then the total amount we need to pay is:

\text{total amount} = 3.75*4.74 = 17.775

The price for 3.75 pounds of cherries is $17.775.

6 0
3 years ago
What is the counter example of when it rains it pours
Ganezh [65]
A counterexample proves something wrong. To disprove "When it rains, it pours," you could give an example of a time when it rains and does not pour. What if it only rains a little? What if it rains frogs? How are you supposed to "pour" frogs? I dunno. This is sort of an open-ended question. I'd go with "It drizzles, but does not pour."
5 0
3 years ago
How do you write 1.45 as a percent?
ICE Princess25 [194]

Answer:

145%

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
LuckyWell [14K]

Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0
3 years ago
There are 3/4 pound of hamburger me there are three hamburger patties that was made each patti weighs the same how much does eac
Luba_88 [7]
1/4 of a pound. 3/4 divided by 3 = 3
7 0
3 years ago
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