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3 years ago

The graphs of functions f(x) and g(x) = f(x) + k are shown:

Mathematics

2 answers:

steposvetlana [31]3 years ago

4 0

It would be D.) K=-2

densk [106]3 years ago

4 0

<h2>Answer:</h2>

The value of k is:

A.) K=2

<h2>Step-by-step explanation:</h2>

We know that the transformation of the type:

f(x) to f(x)+k

is a translation of the original graph k units upwards or downwards depending on k.

if k>0 then the shift is k units up and if k<0 then the shift is k units down.

Here we observe that the graph of the function g(x) is shifted 2 units upwards as compared to the graph of the function f(x).

This means that:

k=2

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The owner of The Pizza Palace has a deal offering 1 free order of breadsticks with 3 pizzas. If 24 total pizzas were sold with t

Sedbober [7]

Answer:

Step-by-step explanation:

for every 3 pizzas 1 free order is served

in 24 pizzas there is 8 (3 pizzas) so 8 free order

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3 years ago

PLEASE HELP!!!!!!!!!!

nydimaria [60]

mere ko question samjh nahi aaya... plz ek baar phir se likho

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3 years ago

Read 2 more answers

Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article "Forces on the Hand in the

Gnoma [55]

Answer:

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

X₂: Force (N) on the hand just after impact on a one- handed backhand drive for an intermediate tennis player.

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!

3 0

3 years ago

Solve x^2+11X<-8 Urgent please help!!!!!

Tamiku [17]

Add 121/4 to each side:
x²+11x+121/4 < 121/4-8
x²+11x+121/4 < 89/4
(x+11/2)² < √89/2 ⇒ -√89/2 < x+11/2 < √89/2
-11/2-√89/2 < x < -11/2+√89/2

8 0

3 years ago

The weight of an object on the moon is

Tanzania [10]

Answer:

40 2/5Ibs or $40\frac{2}{5}$ lbs

Step-by-step explanation:

Let weight of object on the moon be represented = Wm

weight of object on the Earth = We

In the above question, we are told:

The weight of an object on the moon is

5 of its weight on Earth.

Hence, mathematically, this is represented as:

5Wm = We

If a moon rock weighs 202 lbs on Earth

5Wm = 202lbs

Wm = 202lbs/5

Wm = $40\frac{2}{5}$ Ibs

Therefore, the moon rock weighs $40\frac{2}{5}$ Ibs on the moon.

7 0

3 years ago