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sattari [20]
3 years ago
14

Kate bought 4 cans of soup for $2.67. How much would it cost for her to buy 10 cans?

Mathematics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

It would be 6.60

Step-by-step explanation:

first you have to find the price of one can, so you divided 2.67 by 4.

the answer would actually be 0.6675, but you have to stop at the tenths place for money, so in this case, it would be 0.66.

we multiply 0.66 by 10 and get 6.6 which is also 6.60

It would cost $6.60 to but 10 cans.

Idk if this helps, but I hope it does.

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Samantha has 8 quarts of milk.  The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have?   x  a. 1 over 32   x  b. begin mathsize 16px style 1 half end style   x  c. 2   x  d. 32

Step-by-step explanation:Samantha has 8 quarts of milk.  The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have?   x  a. 1 over 32   x  b. begin mathsize 16px style 1 half end style   x  c. 2   x  d. 32Samantha has 8 quarts of milk.  The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have?   x  a. 1 over 32   x  b. begin mathsize 16px style 1 half end style   x  c. 2   x  d. 32Samantha has 8 quarts of milk.  The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have?   x  a. 1 over 32   x  b. begin mathsize 16px style 1 half end style   x  c. 2   x  d. 32

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Step-by-step explanation:

Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

Sample 1: Advanced tennis players

X₁: Force (N) on the hand just after impact on a one- handed backhand drive for an advanced tennis player.

n₁= 6

X[bar]₁= 40.29 N

S₁= 11.29

Sample 2: Intermediate players

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n₂= 8

X[bar]₂= 21.40

S₂= 8.30

Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66

t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

I hope this helps!

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