All categories

3 years ago

Help asap please info in pic

Mathematics

1 answer:

Evgen [1.6K]3 years ago

8 0

U should be all set, u got it right i believe

You might be interested in

Help me!!!

Basile [38]

Answer:

this is your answer look it once.

8 0

3 years ago

The area of a trapezoid is given by the formula A=h/2(a+b) . Solve the formula for b.

Komok [63]

Answer:

(2A)/h -a =b

Step-by-step explanation:

A= [h(a+b)]/2 I rewrote the formula, because h and a+b are in the numerator.

2A=h(a+b) Multiply both sides of equation by 2

(2A)/h= a+b Divide both side by h

(2A)/h -a= b Subtract a from both sides

4 0

3 years ago

Let g(x) = 2x2 - 11x. Find g(-1). A) -13 Eliminate B) -9 C) 0 D) 13

worty [1.4K]

Solution

g(x) = 2 $x^{2}$ - 11x

Plugging in x = -1 in g(x), we get

g(-1) = 2 $(-1)^{2}$ - 11(-1)

Now we have to simplify it.

g(-1) = 2*1 + 11

g(-1) = 2 + 11

g(-1) = 13

So the answer is D) 13.

Thank you. :)

3 0

3 years ago

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)

skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample x⁻ = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

( $(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

$(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )$

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{28-30 }{\frac{10}{\sqrt{25} } }$

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0

3 years ago

Evaluate x(–y + z) for x = 3, y = 3, and z = 1.

Crank

Explanation:
Substitute:

3 for x

3 for y

1 for z

x(−y+z)

becomes:

3(−3+1)⇒

3 ⋅ −2 ⇒

−6

3 0

2 years ago